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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