Lecture 3-1 STUDENT MASTER COPY



Lecture 3-1 STUDENT MASTER COPY

Rigid Bodies: Equivalent Systems of Forces

Rigid Bodies:

1. Usually made up of a group of particles

2. Positions of particles are fixed relative to each other

3. Actual deformations are assumed to be very small

4. Will introduce principle of transmissibility or sliding vectors

5. Will introduce the concept of a moment:

a. moment of a force about a point

b. moment of a force about an axis

6. Will review vector algebra with regard to:

a. vector products of two vectors

b. scalar products of two vectors

7. Will introduce the concept of a couple, i.e. two forces with:

a. equal magnitude

b. parallel lines of action

c. opposite sense

8. Will introduce the concept of a force-couple system:

a. a force acting on a rigid body at a given point can be replaced

b. by a force acting at another point and a couple

Types of Forces Acting on a Rigid Body:

1. External forces:

a. represent the action of other bodies on the given rigid body

b. entirely responsible for the external behavior of the rigid body

c. will cause the rigid body to move or remain in place

d. main subject of Chapters 3, 4, and 5

2. Internal forces:

a. forces which hold together the particles that form the rigid body

b. if the rigid body is structurally composed of several parts, then

the forces that hold the components together are also internal forces

External Forces Acting on a Truck - Figure 3.2 in the text:

1. Weight of the truck:

a. derived from the effect of earth's gravity on the mass of the truck

b. represented by a single force W acting in the negative y direction

c. force W acts at the "center of gravity or mass" of the bus

d. would tend to cause the truck to move downward if not for the ground

2. Reactions R1 and R2:

a. at front and rear axles, respectively:

b. forces R1 and R2 act in the positive y direction

c. the sum of the forces R1 + R2 is equal and opposite to the force W

3. Force F for example:

a. exerted by the men who are pulling the truck

b. point of application is the front bumper where the rope is attached

c. tends to make truck move forward

d. resulting horizontal motion is called a "translation" of the truck

4. Force V for example:

a. exerted by a jack acting upward on the front bumper

b. would cause the center of gravity of the truck to move upward

c. would also cause a "rotation" of the truck about center of gravity

5. All of these external forces acting on the truck which is a rigid body

could cause the truck to translate or rotate or both if unopposed

Principle of Transmissibility:

1. States that the condition of equilibrium or motion

of a rigid body will remain unchanged if:

a. a force F acting at a given point is replaced by a force F’:

i. of the same magnitude and direction

ii. but acting at a different point

b. provided that the two forces have the same line of action

2. The forces F and F’ have the same effect on the rigid body:

a. thus the forces F and F’ are said to be equivalent

b. thus a force F can be "transmitted" along its line of action

3. Based on experimental evidence:

a. can not be derived from the properties presented in this Statics book

b. must be accepted as an "experimental law"

c. will be derived in the textbook for Dynamics

4. Unlike forces applied to particles:

a. forces applied to rigid bodies are not "fixed or bound vectors"

b. forces applied to rigid bodies are free to move along line of action

c. thus forces applied to rigid bodies are "sliding vectors"

External Forces Acting on a Truck - Figure 3.4 in the text:

1. Assuming the force F acts along a line of action passing through the

front and rear bumpers

2. Then F could be replaced by an equivalent force F’:

a. of the same magnitude and direction

b. but acting at the rear bumper

3. the effect on the truck will be the same

4. the forces R1, R2, and W will be unchanged

Rigid Versus Deformable Bodies - Figure 3.5 in the text:

1. Taking a short bar AB with equal and opposite tensile forces P1 and P2:

a. acting at the left and right ends of the bar, respectively

b. by the principle of transmissibility we can replace P2 with P2':

i. where P2' is located at the left end of the bar

ii. thus the forces P1 and P2' will cancel leaving zero force on bar

2. Taking a short bar AB with equal and opposite compressive forces P1 and

P2:

a. acting at the left and right ends of the bar, respectively

b. by the principle of transmissibility we can replace P2 with P2':

i. where P2' is located at the left end of the bar

ii. thus the forces P1 and P2' will cancel leaving zero force on bar

3. Thus if the bar is a rigid bar:

a. there is no difference if P1 and P2 are tensile or compressive forces

b. the bar will react the same

4. If the bar is deformable, however, as all real bars must be, then:

a. the tensile forces will cause:

i. tension in the bar

ii. elongation of the bar

b. the compressive forces will cause:

i. compression in the bar

ii. shortening of the bar

5. Thus the principle of transmissibility should:

a. not be used to determine internal forces, or

b. be used with great care to determine internal forces

Vector Products of Two Vectors - Figures 3.6 to 3.8 in the text:

1. The vector product V or two vectors P and Q has the conditions below:

a. the line of action of V is perpendicular to

the plane defined by the vectors P and Q

b. the magnitude of V is the product of the magnitudes of P and Q and

the sine of the angle Θ formed by P and Q (between 0° and 180°):

i. V = P Q sin Θ

ii. thus the magnitude of V represents the area

of the parallelogram formed by P and Q

c. the sense of V is such that:

i. the vectors P and Q are moved to the same point as the tip of V

ii. a person located at the tip of V will observe as counterclockwise

the rotation through Θ which brings the vector P in line with the

vector Q, i.e. the direction of V follows the right-hand rule

iii. three vectors P, Q, and V form what is called a right-handed triad

2. A vector V which satisfies these three conditions is referred to as the

vector product or cross-product of P and Q which is written as:

V = P X Q

3. If P and Q have the same line of action, then V = 0

4. The vector product V will remain unchanged if we replace Q by another

coplanar vector Q’ such that Q’ begins at the junction of P and Q and

ends along the side of the parallelogram opposite and parallel to P

5. The vector product V is "not commutative", i.e. P X Q does not equal

Q X P, but Q X P = - (P X Q)

Example - Compute the vector product V = P X Q of the vector P with magnitude 6 lying in the xz plane as shown at an angle of 30° from the x axis and the vector Q with magnitude 4 lying along the x axis.

Solution:

V = P Q sin Θ = (6) (4) sin 30° = 12

based on the right-hand rule, the direction of V will be +j, thus:

V = 12 j

Comments on Vector Products:

1. The distributive property does hold true for vector products, thus:

P X (Q1 + Q2) = P X Q1 + P X Q2

2. The associate property does not hold true for vector products, thus:

(P X Q) X S does not equal P X (Q X S)

3. Using the rules for vector products we get the following vector products

for combinations of the unit vectors:

i X i = 0 j X i = -k k X i = j

i X j = k j X j = 0 k X j = -i

i X k = -j j X k = i k X k = 0

4. We can write the vector product V as follows:

V = P X Q = (Px i + Py j + Pz k) X (Qx i + Qy j + Qz k)

5. Applying the distributive law first we get:

V = (Px i) X (Qx i + Qy j + Qz k)

+ (Py j) X (Qx i + Qy j + Qz k)

+ (Pz k) X (Qx i + Qy j + Qz k)

6. Applying the distributive law again we get:

V = (Px i) X (Qx i) + (Px i) X (Qy j) + (Px i) X (Qz k)

+ (Py j) X (Qx i) + (Py j) X (Qy j) + (Py j) X (Qz k)

+ (Pz k) X (Qx i) + (Pz k) X (Qy j) + (Pz k) X (Qz k)

7. Because Θ = 90° between i, j, and k and because we know the vector

product for each combination of unit vectors, we get:

V = 0 + Px Qy k - Px Qz j

- Py Qx k + 0 + Py Qz i

+ Pz Qx j - Pz Qy i + 0

8. Rearranging we get:

V = (Py Qz - Pz Qy) i + (Pz Qx - Px Qz) j + (Px Qy - Py Qx) k

9. Thus, the vector product V can be written as the following

determinant:

V = | i j k |

| Px Py Pz |

| Qx Qy Qz |

Moment of a Force about a Point - Figures 3.12 and 3.13 in the text:

1. Define the force F to be a force applied to a rigid body

2. Define the position vector r of the force F to be a vector connecting a

fixed reference point O with the point of application of the vector F

3. The moment of the force F about the point O is then defined to be the

vector product of r and F as follows: Mo = r X F

4. According to the rules for vector products:

a. the direction of the moment Mo is then defined to be perpendicular

to the plane formed by r and F

b. the sense of Mo follows the right-hand rule turning from the

direction of r to the direction of F

c. the magnitude of Mo is defined to be: Mo = r F sin Θ

5. The distance d is defined as the perpendicular distance from O to the

line of action of F and thus: Mo = F (r sin Θ) = F d

6. The magnitude of the moment Mo is defined as "a measure of the tendency

of the force F to cause the rigid body to rotate about the fixed axis

running along the vector Mo"

a. in SI units the magnitude of Mo is in Newton-meters (N-m)

b. in US units the magnitude of Mo is in pound-feet (lb-ft)

7. The moment Mo does not depend on the point of application of the force

F along its line of action

8. Conversely, the moment Mo can not be used to identify the point of

application of the force F along its line of action

9. If the magnitudes and directions of the moment Mo and the force F are

known, then the line of action of F can also be defined as follows:

a. the line of action of F must lie in a plane through O and

perpendicular to the moment Mo

b. the distance d from O must be equal to the quotient Mo/F of the

magnitudes of Mo and F

c. the sense of Mo determines which side of O the line of action of F

must be drawn

10. The principle of transmissibility states that two forces F and F’ are

equivalent (have the same effect on a rigid body) if they have the

same magnitude, the same direction, and the same line of action.

11. This can now be restated as "two forces F and F’ are equivalent if, and

only if, they are equal (i.e. have the same magnitude and the same

direction) and they have equal moments about a given point O, i.e:

F = F’ and Mo = Mo’

12. For a two-dimensional system in the xy plane:

a. the magnitude of Mo is the magnitude of F times the perpendicular

distance from O to the line of action of F

b. the line of action of Mo will be perpendicular to the xy plane, i.e.

along the z axis

c. the direction of Mo follows the right-hand rule as shown

Varignon's Theorem - Figure 3.14 in the text:

1. The moment Mo of the resultant of several concurrent forces F1, F2, F3,

etc. can be derived using the distributive property as follows:

r X (F1 + F2 + F3 + ...) = r X F1 + r X F2 + r X F3 + ...

2. Thus, "the moment about a given point O of the resultant of several

concurrent forces is equal to the sum of the moments of the various

forces about the same point O"

Rectangular Components of the Moment of a Force - Figures 3.15 and 3.16:

1. Defining the point O as the origin and the point A (x, y, z) as the

point of application of the force F, we can resolve r, F, and Mo into

rectangular components as follows:

r = x i + y j + z k

F = Fx i + Fy j + Fz k

Mo = Mx i + My j + Mz k

2. Since Mo = r X F, we can use the determinant definition of a vector

product to get the following:

Mo = | i j k |

| x y z |

| Fx Fy Fz |

Mx = y Fz - z Fy

My = z Fx - x Fz

Mz = x Fy - y Fx

3. To calculate the moment Mb about an arbitrary point B:

a. we replace the vector r by a vector rba from the point B to the

point of application A of the force F:

Mb = rba X F

b. we define the vectors ra and rb to be vectors from the origin to

points A and B, respectively, thus:

ra = rb + rba

rba = ra - rb = (xa - xb) i + (ya - yb) j + (za - zb) k

= xba i + yba j + zba k

c. using the determinant form of the cross product we get:

Mb = | i j k |

| xba yba zba |

| Fx Fy Fz |

where: xba = xa - xb

yba = ya - yb

zba = za - zb

4. For two-dimensional problems:

a. the magnitude of the moment of a force F about the origin O will be:

Mo = (x Fy - y Fx) k

Mo = Mz = x Fy - y Fx

b. the magnitude of the moment of a force F about an arbitrary point B

will be:

Mb = (xa - xb) Fy - (ya - yb) Fx

Sample Problem 3.1 - A 100-lb vertical force is applied to the end of a lever which has an angle of 60° to the horizontal and is attached to a shaft at point O. Determine:

a) the moment of the 100-lb force about point O

b) the magnitude of the horizontal force applied at point A which would

create the same moment about the point O

c) the smallest force applied at point A which would create the same

moment about the point O

d) how far from the shaft a 240-lb vertical force must act to create the

same moment about the point O

e) whether any one of the forces obtained in parts b), c), and d) is

equivalent to the original force

Solution a):

d = (24 inches) cos 60° = 12 in

Mo = F d = (100 lb) (12 in) = 1200 lb-in

Mo = -(1200 lb-in) k

Solution b):

d = (24 inches) sin 60° = 20.78 in

Mo = 1200 lb-in = F d = F (20.78 in)

F = (1200 lb-in) / (20.78 in) = 57.75 lb

F = (57.75 lb) i

Solution c):

d = 24 inches

Θ = -30°

Mo = 1200 lb-in = F d = F (24 in)

F = (1200 lb-in) / (24 in) = 50 lb

Solution d):

Mo = 1200 lb-in = F d = (240 lb) d

d = (1200 lb-in) / (240 lb) = 5 inches

OB = d / cos 60° = (5 inches) / (0.50) = 10 inches

Solution e):

No because all of the force have different magnitudes (100 lb, 57.75 lb,

50 lb, and 240 lb)

Sample Problem 3.2 - A force of 800 N acts on a bracket as shown.

Determine the moment of the force about point B.

Solution:

Mba = rba X F

rba = - (0.2 m) i + (0.16 m) j

F = (800 N) cos 60° i + (800 N) sin 60° j

= (400.0 N) i + (692.8 N) j

Mb = (xa - xb) Fy - (ya - yb) Fx

= -(0.2 m) (692.8 N) - (0.16 m) (400.0 N) = -202.6 N-m

Mb = -(202.6 N-m) k

Sample Problem 3.3 - A 30-lb force acts on the end of the 3-ft lever as shown. Determine the moment of the force about the point O.

Solution:

Mb = roa X F

roa = (3 ft) cos 50° i + (3 ft) sin 50° j

= (1.929 ft) i + (2.298 ft) j

Θ = 50° - 20° = 30°

F = (30 lb) cos 30° i + (30 lb) sin 30° j

= (25.98 lb) i + (15.00 lb) j

Mb = (xa - xb) Fy - (ya - yb) Fx

= (1.929 ft) (15.00 lb) - (2.298 ft) (25.98 lb) = -30.77 lb-ft

Mb = -(30.77 lb-ft) k

Sample Problem 3.4 - A rectangular plate is supported by brackets at

points A and B and by a wire CD between points C and D as shown. If the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at point C.

Solution:

Ma = rac X F

rac = (0.300 m) i + (0.080 m) k

F = F la = (200 N) CD / CD

CD = -(0.300 m) i + (0.240 m) j - (0.320 m) k

CD = √ [(-0.300 m)² + (0.240 m)² + (-0.320 m)²] = 0.500 m

F = (200 N) [-(0.300 m) i + (0.240 m) j - (0.320 m) k] / 0.500 m

= -(120 N) i + (96 N) j - (128 N) k

Ma = (ry Fz - rz Fy) i + (rz Fx - rx Fz) j + (rx Fy - ry Fx) k

= [(0.000 m) (-128 N) - (0.080 m) (96 N)] i

+ [(0.080 m) (-120 N) - (0.300 m) (-128 N)] j

+ [(0.300 m) (96 N) - (0.000 m) (-120 N)] k

= -(7.68 N-m) i + (28.8 N-m) j + (28.8 N-m) k

Ma = √ [(-7.68 N-m)² + (28.8 N-m)² + (28.8 N-m)²] = 41.45 N-m

Lecture 3-2 - Scalar Product of Two Vectors STUDENT MASTER COPY

Scalar Product of Two Vectors:

1. The scalar product of two vectors P and Q is defined as:

a. the product of the magnitude of P and Q and the cosine of the angle

Θ formed by P and Q which is written as: P ⋅ Q = P Q cos Θ

b. the product is defined as a scalar and not as a vector

c. the product is also referred to as the dot product

2. The scalar product of two vectors P and Q is commutative:

P ⋅ Q = Q ⋅ P

3. The scalar product of two vectors P and Q is distributive:

P ⋅ (Q1 + Q2) = P ⋅ Q1 + P ⋅ Q2

4. The associate law does not apply to the scalar product of two vectors

5. The dot products of the unit vectors will be as follows:

i ⋅ i = (1) (1) cos 0° = 1

j ⋅ j = (1) (1) cos 0° = 1

k ⋅ k = (1) (1) cos 0° = 1

i ⋅ j = (1) (1) cos 90° = 0

j ⋅ k = (1) (1) cos 90° = 0

k ⋅ i = (1) (1) cos 90° = 0

6. Using the distributive law and dropping out the terms with unit vector

dot products that are zero we get:

P ⋅ Q = (Px i + Py j + Pz k) ⋅ (Qx i + Qy j + Qz k)

= (Px i) ⋅ (Qx i) + (Py j) ⋅ (Qy j) + (Pz k) ⋅ (Qz k)

= Px Qx + Py Qy + Pz Qz

7. Thus, the dot product of the same vector P ⋅ P will be:

P ⋅ P = Px² + Py² + Pz²

Applications of Scalar Products or Dot Products:

1. Angle formed by two given vectors:

a. given two vectors P and Q, their dot product can be written as:

P ⋅ Q = P Q cos Θ = Px Qx + Py Qy + Pz Qz

b. thus we can write:

Θ = arccos [(Px Qx + Py Qy + Pz Qz) / (P Q)]

2. Projection of a vector on a given axis - Figures 3.21 to 3.23 in text:

a. given a vector P with an angle Θ to a given line OL, the projection

of P on the axis OL will be a scalar:

Pol = P cos Θ

b. given a vector Q acting along the line OL, then the scalar product

P ⋅ Q will be:

P ⋅ Q = P Q cos Θ = Pol Q

c. thus we can write:

Pol = P ⋅ Q / Q = (Px Qx + Py Qy + Pz Qz) / Q

d. if Q is the unit vector la along OL, then:

Pol = P ⋅ Q / Q = P ⋅ la / la = P ⋅ la / 1 = P ⋅ la

e. since la = (cos Θx) i + (cos Θy) j + (cos Θz) k:

Pol = Px cos Θx + Py cos Θy + Pz cos Θz

Mixed Triple Product of Three Vectors:

1. We define the mixed triple product of three vectors S, P, and Q to be:

S ⋅(P X Q)

2. Thus the triple product of the vectors S, P, and Q is formed by taking

the scalar product of the vector S with the vector product of P and Q

3. The geometric interpretation for the mixed triple product of three

vectors is as follows - Figures 3.24 and 3.25:

a. the magnitude of V = P X Q is the area of the parallelogram formed

by the vectors P and Q

b. the vector V = P X Q is perpendicular to the

plane formed by the vectors P and Q

c. the scalar product S ⋅ V = S V cos Θ = V S cos Θ

d. the projection of S onto the line of action of V which is

S cos Θ = Sol

e. thus S ⋅ V = (height Sol) (area of parallelogram)

f. therefore S ⋅ (P X Q) represents the area of the parallelepiped

formed by the three vectors S, P, and Q

4. Rules governing the sign of the mixed triple product are as follows:

a. S ⋅ (P X Q) = P ⋅ (Q X S) = Q ⋅ (S X P)

= - S ⋅ (Q X P) = - P ⋅ (S X Q) = - Q ⋅ (P X S)

b. if the order is an alphabetical or circular permutation,

i.e. S to P to Q to S to P to Q, etc., then the sign will be the

same as S ⋅ (P X Q)

c. if the order is a reverse alphabetical or circular permutation, i.e.

Q to P to S to Q to P to S, etc. then the sign will be the opposite

of S ⋅ (P X Q)

5. The rectangular components of the mixed triple product are as follows:

a. S ⋅ (P X Q) = S ⋅ V = Sx Vx + Sy Vy + Sz Vz

b. P X Q = (Py Qz - Pz Qy) i + (Pz Qx - Px Qz) j + (Px Qy - Py Qx) k

c. S ⋅ (P X Q) = Sx (Py Qz - Pz Qy) + Sy (Pz Qx - Px Qz)

+ Sz (Px Qy - Py Qx)

6. Thus, the mixed triple product can be written as the determinant:

S ⋅ (P X Q) = | Sx Sy Sz |

| Px Py Pz |

| Qx Qy Qz |

Moment of a Force about a Given Axis - Figures 3.27 and 3.28:

1. Given a force F acting on a rigid body and the moment Mo of that force

acting at the origin O

2. Let OL be an arbitrary line acting through the origin O

3. The moment of F about the line OL is defined as the projection of the

moment Mo on to the line OL

4. We know the following is true:

Mol = Mo ⋅ la = la ⋅ Mo

Mo = (r X F)

Mol = la ⋅ (r X F)

5. Thus, the magnitude of the moment about the line OL is formed by the

mixed triple product of the vectors la, r, and F:

6. Using the determinant form of the mixed triple product we get:

Mol = la ⋅ (r X F) = | lax lay laz |

| x y z |

| Fx Fy Fz |

where: lax, lay, laz = cos Θx, cos Θy, cos Θz, respectively

= the direction cosines of the line OL

x, y, z = coordinates of the point of application of force F

Fx, Fy, Fz = the components of the force F

7. To further define the physical significance of Mol:

a. resolve F into two components F1 and F2 where F1 is parallel to the

line OL and F2 lies in a plane that is perpendicular to the line OL

b. resolve r into two components r1 and r2 where r1 is parallel to the

line OL and r2 lies in a plane that is perpendicular to the line OL

c. by substitution we get: Mol = la ⋅ (r1 + r2) X (F1 + F2)

= la ⋅ r1 X F1 + la ⋅ r1 X F2

+ la ⋅ r2 X F2 + la ⋅ r1 X F2

d. la ⋅ r1 X F1 = 0 because the vectors r1 and F1 are parallel

e. la ⋅ r1 X F2 = 0 because the cross product r1 X F2 lies in the plane

P that is perpendicular to la

f. la ⋅ r2 X F1 = 0 because the cross product r2 X F1 also lies in the

plane P that is perpendicular to la

g. thus we get: Mol = la ⋅ r2 X F2

h. the vector product r2 X F2 is perpendicular to the plane P and

represents the moment of the component F2 of F about the point Q

where the line OL intercepts the plane P

i. Mol is positive if r2 X F2 has the same direction as OL and negative

if r2 X F2 has the opposite direction as OL

j. thus Mol measures the tendency of F2 to make the rigid body rotate

about the fixed axis OL

k. since F1 contributes nothing to Mol, we can conclude that the moment

Mol of the force F measures the tendency of F to make the rigid

body rotate about the fixed axis OL

8. Given a force F applied at a point A, the moment of the force about a

line BL that does not pass through the origin will be:

a. Mbl = la ⋅ Mb = la ⋅ (rba X F)

where: rba is a vector from a point B on the line BL to the point A

b. since rb + rba = ra

rba = ra - rb = (xa - xb) i + (ya - yb) j + (za - zb) k

= xba i + yba j + zba k

c. then the determinant form of Mbl will be:

Mbl = la ⋅ (rba X F) = | lax lay laz |

| xba yba zba |

| Fx Fy Fz |

d. the final result is independent of the location of the point B, if

we chose another point C, then we would get:

rca = ra - rc

Mcl = la ⋅ [(la -rc) X F]

= la ⋅ [(ra - rb) X F] + la ⋅ [(rb - rc) X F]

e. since (rb - rc) and la are colinear, la will lie in the same plane

as the cross product (rb - rb) X F and thus the mixed triple

product will be zero

Sample Problem 3.5 - A cube of side "a" is acted upon by a force P as

shown. Determine the moment of P:

A) about A

B) about the edge AB

C) about the diagonal AG

D) using the result of Solution C), determine the perpendicular distance

from AG to FC

Solution A):

raf = a i - a j = a (i - j)

P = (P / √2) j - (P / √2) k = (P / √2) (j - k)

Ma = raf X P = a (i - j) X (P / √2) (j - k)

= (a P / √2) [i X j - i X k - j X j + j X k)]

= (a P / √2) [k + j + 0 + i)] = (a P / √2) [i + j + k]

Solution B):

Mab = i ⋅ Ma = i ⋅ (a P / √2) [i + j + k]

= (a P / √2) [i ⋅ i + i ⋅ j + i ⋅ k] = (a P / √2) [1 + 0 + 0] = a P / √2

Solution C) - Method #1:

la = AG / AG = [a i - a j - a k] / a √3 = (1 / √3) [i - j - k]

Mag = la ⋅ Ma = (1 / √3) [i - j - k] ⋅ (a P / √2) [i + j + k]

= (a P / √6) [i ⋅ i + i ⋅ j + i ⋅ k - j ⋅ i - j ⋅ j - j ⋅ k - k ⋅ i - k ⋅ j

- k ⋅ k]

= (a P / √6) [1 + 0 + 0 + 0 - 1 + 0 + 0 + 0 - 1] = - a P / √6

Solution C) - Method #2:

Mag = | lax lay laz | = | (1 / √3) (-1 / √3) (-1 / √3) |

| xaf yaf zaf | | a -a 0 |

| Px Py Pz | | 0 (P / √2) (-P / √2) |

= (1 / √3) (- a) (- P / √2) + (1 / √3) (a) (- P / √2)

+ (-1 / √3) (a) (P / √2) = - a P / √6

Solution D):

Since the force P lies along the line FC, the perpendicular distance "d"

from the diagonal AG to the line FC is also the distance from the line AG

to the line of action of the force P. Thus the magnitude of the moment

Mag can be expressed as: Mag = - P d = - a P / √6

dividing both sides by -P we get: d = a / √6

Moment of a Couple - Figures 3.31 to 3.33 in the text:

1. Two forces F and -F having the same magnitude, parallel lines of

action, and opposite sense are said to form a couple.

2. The sum of the components of the two forces in any direction will be

zero.

3. The sum of the moments of the two forces about a given point is not

zero.

4. Thus the two forces will tend to rotate any rigid body that they act

on.

5. Denoting ra and rb as the position vectors for the points of

application of the two forces F and -F, we find the sum of the moments

of the two forces about the origin O as follows:

ra X F + rb X -F = (ra - rb) X F

6. Defining r to be the vector joining the points of application

B and A of the forces -F and F, respectively, we get:

r = ra - rb

(ra - rb) X F = r X F = M

where M is called the moment of the couple (i.e. "every couple has

its moment!")

7. The magnitude of the moment M will be:

M = r F sin Θ = F (r sin Θ) = F d

where d is the perpendicular distance between the lines of action of

the forces F and -F

8. Since F and d are independent of the origin O, the moment M

can be calculate about any point O without changing the final result

9. Thus the moment M is a free vector which may be applied at any point

10. Two couples formed by pairs of equal and opposite forces F1 & -F1 and

F2 & -F2 with perpendicular distances of d1 and d2, respectively,

between their lines of action will have equal moments if all of the

following are true:

a. F1 d1 = F2 d2

b. the two couples lie in the same plane

c. the two couples have the same sense (both clockwise or

both counterclockwise)

Equivalent Couples - Figure 3.34 in the text:

1. Consider the three couples shown in Figure 3.34 which will act

successively on the same rectangular box with x, y, z dimensions of 6,

4, and 4 inches

2. Each couple would impart a rotation about the y axis

3. Each couple would have the same moment of 120 lb-in or

M = (120 lb-in) j thus they would seem to be equivalent

4. We can prove that two systems of forces are equivalent (i.e. they have

the same effect on a rigid body) if we can transform one system into

the other by means of one or more of the following operations:

a. replacing two forces acting on the same particle by their resultant

b. resolving a force into two components

c. canceling two equal and opposite forces acting on the same particle

d. attaching to the same particle two equal and opposite forces

e. moving a force along its line of action

5. Each of the above operations can be justified based on the following:

a. the parallelogram law

b. the principle of transmissibility

Proving that Two Couples in the Same Plane are Equivalent - Figure 3.35:

1. Given two couples in the same plane F1 & -F1 and F2 & -F2 with

magnitudes of F1 and F2, respectively, and with perpendicular

distances of d1 and d2, respectively, between their lines of action

2. Since both couples have the same moment M: F1 d1 = F2 d2

3. Denoting points A, B, C, and D as the intersection points

of the four lines of action of the four forces as shown

4. Slide forces F1 and -F1 so they are applied at points A and B,

respectively

5. Resolve F1 into two components P and Q along the lines BA and AC,

respectively

6. Resolve -F1 into two components -P and -Q along the lines AB and BD,

respectively

7. The forces P and -P have the same magnitude, the same line of action,

and the opposite direction, thus P can be slid along the line of action

and applied at point B, for example, and then P and -P will cancel out

8. Thus the couple F1 and -F1 is transformed into a couple formed by Q

and -Q

9. The moment of Q and -Q can then be derived with respect to point B

10. The moment of F1 and -F1 can then be derived with respect to point B

11. By Varignon's Theorem, the moment of F1 with respect to point B is

equal to the sum of the moments of the components P and Q

12. The moment of P with respect to point B is zero, thus:

M = F1 d1 = Q d2

M = F2 d2

Q = F2

13. Thus the forces Q and -Q are equal to the forces -F2 and F2,

respectively, and the couples F1 & -F1 and F2 & -F2 are equivalent

Proving that Two Couples in Parallel Planes are Equivalent - Figure 3.36:

1. Given two couples (F1 & -F1 and F2 & -F2) in parallel planes P1 and P2,

respectively, with the same magnitudes F and the same perpendicular

distances d

2. Thus, we will have all four forces acting along parallel lines of

action as shown in Figures 3.36a) and 3.36d)

3. Define a plane P3 that contains -F1 and F2 and a plane P4 that contains

F1 and -F2

4. The intersection of P3 and P4 will be a line parallel to the lines of

action of the four forces and halfway between planes P1 and P2 as shown

5. Along this line of intersection we apply two equal and opposite forces

F3 and -F3 with magnitudes of F as shown in Figure 3.36b)

6. The couple formed by F1 and -F3 has the same moment and lies in the

same plane as the couple F3 and -F2

7. The couple formed by -F1 and F3 has the same moment and lies in the

same plane as the couple -F3 and F2

8. Thus we replace the couples F1 & -F3 and -F1 and F3 by the couples

F3 & -F2 and -F3 & F2, respectively, as shown in Figure 3.36c)

9. Canceling the forces F3 and -F3, we are left with the couple F2 and

-F2 which must therefore be equivalent to the couple F1 and -F1

Addition of Couples - Figure 3.37 in the text:

1. Consider two intersecting planes P1 and P2 and with couples

F1 & -F1 and F2 & -F2, respectively, and with moments M1 and M2,

respectively,

2. Assume the forces F1 and F2 act at point A and the forces -F1 and F2

act at point B, where the line of action AB is perpendicular to all

four forces

3. The resultant R of F1 + F2 will form a couple with the resultant -R

of -F1 + -F2

4. Denoting r as the vector from B to A and using the definition for the

moment M of a couple we get: M = r X R = r X (F1 + F2)

5. By Varignon's Theorem we get: M = r X F1 + r X F2 = M1 + M2

Representing Couples by Couple Vectors - Figure 3.38 in the text:

1. A couple can be represented by the vector M of its moment with the

addition of a circular arrow to denote a couple vector, i.e. a moment

vector formed by a couple as shown in Figure 3.38b)

2. Like a moment, couple vectors are free vectors

3. Also like a moment, a couple vector can be resolved into rectangular

components as shown in Figure 3.38d)

4. Thus the x, y, and z components represent couples acting in the yz, xz,

and xy planes, respectively

Resolution of a Given Force into a Force at O and a Couple - Figure 3.39:

1. Consider a force F acting on a rigid body at a point A defined by the

position vector r as shown in Figure 3.39a)

2. To move the force to point O we add two vectors F and -F at the origin O

as shown in Figure 3.39b)

3. The force F at point A and the force -F at the origin O now form a

couple represented by the couple vector Mo as shown in Figure 3.39c)

4. Thus the force F at A has been transformed into a force F at the origin

and a couple Mo

5. For a different point O' with a vector s from O' to O we would get:

r’ = s + r

Mo’ = r’ X F = (s + r) X F = s X F + r X F = s X F + Mo

6. Thus, the moment Mo’ of F about O' is obtained by adding the vector

product s X F to the moment Mo of F about the origin O

7. Conversely, given a couple vector Mo and a force F that lies in a plane

perpendicular to Mo, then they can be replaced by moving the force F in

the plane perpendicular to Mo until its moment about O is equal to the

moment Mo of the couple to be replaced

Sample Problem 3.6 - Determine the components of the single couple

equivalent to the two couples shown.

Solution #1:

The calculations can be made much easier by applying equal and opposite

forces of 20 lb at point A, then:

Mx = - (30 lb) (18 inches) = -540 lb-in

My = (20 lb) (12 inches) = 240 lb-in

Mz = (20 lb) (9 inches) = 180 lb-in

M = - (540 lb-in) i + (240 lb-in) j + (180 lb-in) k

M = 617.7 lb-in

Solution #2:

The problem can also be solved by taking moments about any point.

The easiest would be point D which has two forces applied. Then:

M = Md = r1 X F1 + r2 X F2

= (18 in) j X (-30 lb) k + [(9 in) j - (12 in) k] X (-20 lb) i

= - (540 lb-in) j X k - (180 lb-in) j X i + (240 lb-in) k X i

= - (540 lb-in) i + (180 lb-in) k + (240 lb-in) j

= - (540 lb-in) i + (240 lb-in) j + (180 lb-in) k

M = 617.7 lb-in

Sample Problem 3.7 - Replace the couple and force shown by an equivalent

single force applied to the lever. Determine the distance from the shaft

to the point of application of this equivalent force.

Solution #1:

Step #1: Move force F to origin and calculate moment Mo due to F:

Mo = OB X F = [(0.150 m) i + (0.260 m) j] X - (400 N) j

= - (60 N-m) i X j - (104 N-m) j X j

= - (60 N-m) k

Step #2: Calculate total moment M at the origin:

M = Mo + (0.120 m) j X (200 N) i = - (60 N-m) k + (24 N-m) j X i

= - (60 N-m) k - (24 N-m) k = - (84 N-m) k

Step #3: Move force F to point C so that moment generated will equal M:

M = - (84 N-m) k = OC X F

= [(OC cos 60°) i + (OC sin 60°) j] X - (400 N) j

= - [(200 N) OC] i X j - [(346.4 N) OC] j X j

= - [(200 N) OC] k

- 84 N-m = - (200 N) OC

OC = (-84 N-m) / (-200 N) = 0.420 m = 420 mm

Solution #2:

Step #1: Move couple to point B and calculate moment Mo due to couple:

M = (0.120 m) j X (200 N) i = (24 N-m) j X i = - (24 N-m) k

Step #2: Move force F to point C so that moment generated will equal M:

M = - (24 N-m) k = BC X F

= [(BC cos 60°) i + (BC sin 60°) j] X - (400 N) j

= - [(200 N) BC] i X j - [(346.4 N) BC] j X j

= - [(200 N) BC] k

- 24 N-m = - (200 N) BC

BC = (-24 N-m) / (-200 N) = 0.120 m = 120 mm

Step #3: Add OB and BC to get OC:

OC = OB + BC = 300 mm + 120 mm = 420 mm

Lecture 3-3 STUDENT MASTER COPY

Systems of Forces Acting on a Rigid Body:

Reduction of a System of Forces to One Force and One Couple - Figures 3.41 and 3.42:

1. Consider a system of forces F1, F2, F3, etc., acting on a rigid body at

points A1, A2, A3, etc., defined by the position vectors r1, r2, r3,

etc. as shown in Figure 3.41a)

2. The force F1 can be moved from A1 to a given point O if a couple of

moment M1 equal to the moment r1 X F1 of F1 about point O is added to

the original system of forces

3. The same process can be applied to forces F2, F3, etc., and thus we get

the system shown in Figure 3.41b) consisting of forces and couples

acting at point O

4. Since the new forces and moments are concurrent, they can be added

vectorially and replaced by the resultant force R and the moment

resultant or couple vector Mor

5. Thus, any system of forces no matter how complex can be reduced to an

equivalent force-couple system acting at a given point O

6. While each couple vector M1, M2, M3, etc. is perpendicular to the

corresponding forces F1, F2, F3, etc., the resultant force R is not in

general perpendicular to the resultant couple vector Mor

7. The equivalent force-couple system is defined by the equations:

R = Σ F

Mor = Σ Mo = Σ r X F

8. Once a system of forces has been reduced to a force-couple system of a

resultant moment Mor and a resultant force R at a point O, it can be

reduced to a force-couple system at another point O' as follows:

Mor’ = Mor + s X R

where s is the vector from point O' to point O

9. The resultants R and Mor can be written in rectangular coordinates as:

R = Rx i + Ry j + Rz k

Mor = Mox i + Moy j + Moz k

where: Rx, Ry, and Rx measure the tendency of the resultant force R to

move the rigid body in the x, y, and z directions, respectively

Mox, Moy, and Moz measure the tendency of the resultant moment

Mor to rotate the rigid body about the x, y, and z axes,

respectively

Equivalent Systems of Forces Acting on a Rigid Body:

1. Two systems of forces acting on a rigid body are equivalent if they may

be reduced to the same force-couple system at a given point O

2. Two systems of forces F1, F2, F3, etc. and F1', F2', F3', etc., both

acting on the same rigid body are equivalent if, and only if, the sums

of the forces and the sums of the moments about a given point O of the

forces of the two systems are, respectively, equal

3. This can be written mathematically in vector form as follows:

Σ F = Σ F’

Σ Mo = Σ Mo’

4. This can be written mathematically in scalar form as follows:

Σ Fx = Σ Fx' Σ Fy = Σ Fy' Σ Fz = Σ Fz'

Σ Mx = Σ Mx' Σ My = Σ My' Σ Mz = Σ Mz'

Equipollent Systems of Vectors:

1. If two systems of forces acting on a rigid body are equipollent, they

are also equivalent

2. The difference between equipollent and equivalent is:

a. equivalent systems are applied to the same rigid body

b. equipollent systems may be applied to a series of particles and thus

the effect on any one particle may not be the same, i.e if two

systems of forces acting on a series of particles are equipollent,

they may not be equivalent

Reduction of a System of Forces to a Single Couple or a Single Force:

1. If the resultant force R of a system of forces applied to a rigid body

is zero, then the system can be reduced to a single moment couple Mor,

which is called the resultant couple of the system

2. A system of forces acting on a rigid body may be reduced to a single

resultant force R if the resultant moment Mor with respect to a point O

and the resultant force R are mutually perpendicular

3. For a system of forces applied to a rigid body, the resultant moment

Mor and the resultant force R will be perpendicular if one of the

following are true:

a. the system of forces is concurrent (applied at the same point or

particle), then they can be added with no resultant moment Mor

b. the system of forces is coplanar (acts in the same plane)

c. the forces in the system have parallel lines of action

Systems of Coplanar Forces in the x-y Plane - Figures 3.43 and 3.44:

1. The sum R of the forces of the system will also lie in the same plane,

while the moment of each force about the origin O, and thus the moment

resultant Mor about the origin O, will be perpendicular to the plane

2. The force-couple system therefore consists of a resultant force R and

a resultant moment Mo that are mutually perpendicular (Figure 3.43b)

3. The force R and the moment Mo can be replaced by the single force R by

moving R in the plane until its moment about the origin O is equal to

Mo (Figure 3.43c)

4. To derive the formula for the correct line of action for R (Figure

3.43c):

a. resolve the force R into rectangular components (Figure 3.44a):

R = Rx i + Ry j

b. move R along the x axis to the point B (x1, 0), where the following

is true as shown in Figure 3.44b):

x1 = Mor / Ry

because the line of action of the x component Rx of R passes

through the origin and thus has not no moment contribution

c. move R along the y axis to the point C (0, y2), where the following

is true as shown in Figure 3.44c): y2 = - Mor / Rx

because the line of action of the y component Ry of R passes

through the origin and thus has not no moment contribution

d. given two points B and C on the line of action, we can derive the

formula for the line as follows:

slope from C to B = δy / δx = (y1 - y2) / (x1 - x2)

= [0 - (- Mor / Rx)] / [(Mor / Ry) - 0] = Ry / Rx

y intercept = y2 = - Mor / Rx

y = - (Mor / Rx) + (Ry / Rx) x

e. as a check, assume x = 0, then:

y = - Mor / Rx

f. as a second check, if y = 0, then:

0 = - (Mor / Rx) + (Ry / Rx) x

x = (Mor / Rx) / (Ry / Rx) = Mor / Ry

Systems of Forces Parallel to the y Axis - Figure 3.45 in the text:

1. Since all forces are parallel to the y axis, the resultant force R will

also be parallel to the y axis (as shown in Figures 3.45a and 3.45b)

2. Since the moment of each force must be perpendicular to the force, the

resultant moment Mor will lie in the xz plane with no y component (as

shown in Figure 3.45b)

3. The force-couple system acting at the origin will consist of the

resultant force R and the resultant moment Mor which are mutually

perpendicular (as shown in Figure 3.45b)

4. Because the resultant moment Mor lies in the xz plane, it can be

resolved into two components Mxr and Mzr (as shown in Figure 3.45b)

5. The force R and the moment Mor can be replaced by the single force R by

moving R to a point A (x, 0, z) such that its moment about the origin O

is equal to Mor (Figure 3.45c):

Mor = Mxr i + Mzr k = r X R = (x i + z k) X (Ry j)

= x Ry i X j + z Ry k X j = x Ry k - z Ry i

Mzr = x Ry

x = Mzr / Ry = Mzr / R

Mxr = - z Ry

z = - Mxr / Ry = - Mxr / R

Sample Problem 3.8 - A 4.80-meter beam is subjected to the forces shown.

Reduce the given system of forces to:

a) an equivalent force-couple system at point A

b) an equivalent force-couple system at point B

c) a single resultant force

Solution a):

R = Σ F = (150 N) j - (600 N) j + (100 N) j - (250 N) j = - (600 N) j

Mar = Σ r X F = (1.6 m) i X (-600 N) j + (2.8 m) i X (100 N) j

+ (4.8 m) i X (-250 N) j

= - (960 N-m) k + (280 N-m) k - (1200 N-m) k = - (1880 N-m) k

Solution b):

R = - (600 N) j

Mbr = Mar + BA X R = - (1880 N-m) k + (-4.8 m) i X (-600 N) j

= - (1880 N-m) k + (2880 N-m) k = (1000 N-m) k

Solution c):

Mcr = Mar + EA X R = - (1880 N-m) k + (-x) i X (-600 N) j

= - (1880 N-m) k + (600 x N) k = 0

x = 1880 N-m / 600 N = 3.133 m

Sample Problem 3.9 - Four tugboats are used to bring an ocean liner into port. Each tugboat exerts a 5000-lb force in the direction and at the location shown. Determine:

a) the equivalent force-couple system at the foremast O

b) the point on the hull where a single, more powerful tugboat should

push to produce the same effect as the original four tugboats

Solution a):

R = Σ F = (5000 lb cos 60°) i - (5000 lb sin 60°) j

+ (3000 lb) i - (4000 lb) j - (5000 lb) j

+ (5000 lb cos 45°) i + (5000 lb sin 45°) j

= (2500 lb + 3000 lb + 3536 lb) i

+ (-4330 lb - 4000 lb - 5000 lb + 3536 lb) j

= (9036 lb) i - (9794 lb) j

Mor = Σ r X F = [(-90 ft) i + (50 ft) j] X [(2500 lb) i - (4330 lb) j]

+ [(100 ft) i + (70 ft) j] X [(3000 lb) i - (4000 lb) j]

+ [(400 ft) i + (70 ft) j] X [- (5000 lb) j]

+ [(300 ft) i - (70 ft) j] X [(3536 lb) i + (3536 lb) j]

= (389,700 ft-lb) k - (125,000 ft-lb) k - (400,000 ft-lb) k

- (210,000 ft-lb) k - (2,000,000 ft-lb) k

+ (1,061,000 ft-lb) k + (247,500 ft-lb) k

= -(1,037,000 ft-lb) k

R = √[(9036 lb)² + (-9794 lb)²] = 13,330 lb

Θ = arctan (-9794 lb / 9036 lb) = - 47.31°

Solution b):

y = - (Mor / Rx) + (Ry / Rx) x

= - (-1,037,000 ft-lb / 9036 lb) + (-9794 lb / 9036 lb) x

= 114.8 ft - 1.084 x

Since Ry < 0, the bigger tugboat must be pushing the

ocean liner along the starboard side, thus y = 70 ft and:

70 ft = 114.8 ft - 1.084 x

x = (70 ft - 114.8 ft) / (-1.084) = 41.33 ft

A = (x, y) = (41.33 ft, 70 ft)

Sample Problem 3.10 - Three cables are attached to a bracket as shown. Replace the forces exerted by the cables by an equivalent force-couple system at point A.

Solution:

labe = BE / BE = BE / √[(0.075 m)² + (-0.150 m)² + (0.050 m)²]

= (0.075 m i - 0.150 m j + 0.050 m k) / (0.175 m)

rab = 0.075 m i + 0.050 m k

Fb = Fb labe = (700 N / 0.175 m) (0.075 m i - 0.150 m j + 0.050 m k)

= 300 N i - 600 N j + 200 N k

rac = 0.075 m i - 0.050 m k

Fc = (1000 N) (cos 45°) i - (1000 N) (sin 45°) k = 707.1 N i - 707.1 N k

rad = 0.100 m i - 0.100 m j

Fd = (1200 N) (cos 60°) i + (1200 N) (sin 60°) j = 600 N i + 1039 N j

R = (300 N + 707.1 N + 600 N) i + (- 600 N + 1039 N) j

+ (200 N - 707.1 N) k = (1607 N) i + (439.0 N) j - (507.1 N) k

rab X Fb = | i j k | = 30.00 N-m i - 45.00 N-m k

| 0.075 m 0.000 m 0.050 m |

| 300 N -600 N 200 N |

rac X Fc = | i j k | = 17.68 N-m j

| 0.075 m 0.000 m -0.050 m |

| 707 N 0 N -707 N |

rad X Fd = | i j k | = 163.9 N-m k

| 0.100 m -0.100 m 0.000 m |

| 600 N 1039 N 0 N |

Mar = (30.00 N-m) i + (17.68 N-m) j + (118.9 N-m) k

Sample Problem 3.11 - A square foundation mat supports the four columns

shown. Determine the magnitude and point of application of the resultant of the four loads.

Solution:

ro = 0

ra = (10 ft) i

rb = (10 ft) i + (5 ft) k

rc = (4 ft) i + (10 ft) k

Fo = - (40 k) j

Fa = - (12 k) j

Fb = - (8 k) j

Fc = - (20 k) j

ro X Fo = 0

ra X Fa = - (120 k-ft) k

rb X Fb = - (80 k-ft) k + (40 k-ft) i

rc X Fc = - (80 k-ft) k + (200 k-ft) i

R = - (80 k) j

Mor = (240 k-ft) i - (280 k-ft) k

r = (x i + z k) = vector from origin to point of application of R

r X R = [x i + z k] X [- (80 k) j]

= - (80 k) (x) k + (80 k) (z) i

= (240 k-ft) i - (280 k-ft) k

x = - 280 k-ft / - 80 k = 3.500 ft

(80 k) z = 240 k-ft

z = 240 k-ft / 80 k = 3.000 ft

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