Lesson 11: Equilibrium, Newton’s second law, Rolling, Angular Momentum ...
Lesson 11: Equilibrium, Newton's second law, Rolling, Angular Momentum (Sections 8.38.9)
Last time we began discussing rotational dynamics. We showed that the rotational inertia depends on the shape of the object and the location of the rotational axis. We also learned that the torque changes the state of rotation.
Rotational Equilibrium
We remember that for an object to remain at rest, the net force acting on it must be equal to zero. (Newton's first law.) However, that condition is not
F2
sufficient for rotational equilibrium. What happens
to the object to the right? The forces have the same
magnitude.
F1 F2
F1
Conditions for equilibrium (both translational and rotational):
F 0 and 0
The obedient spool. F1 and F2 make the spool roll to the left, F4 to the right, and F3 makes it slide.
Problem-Solving Steps in Equilibrium Problems (page 274) 1. Identify an object or system in equilibrium. Draw a diagram showing all the forces acting on that object, each drawn at its point of application. Use the center of gravity (CM) as the point of application of any gravitational forces. 2. To apply the force conditions, choose a convenient coordinate system and resolve each force into its x- and y-components. 3. To apply the torque condition, choose a convenient rotation axis ? generally one that passes through the point of application of an unknown force. Then find the torque due to each force. Use whichever method is easier: either the lever arm times the magnitude of the force or the distance times the perpendicular component of the force. Determine the direction of each torque; then either set the sum of all torques (with their algebraic signs)
Lesson 11, page 1
Lesson 11: Equilibrium, Newton's second law, Rolling, Angular Momentum (Sections 8.38.9)
equal to zero or set the magnitude of the CW torques equal to the magnitudes of the CCW torques. 4. Not all problems require all three equilibrium equations (two force component equations and one torque equation). Sometimes it is easier to use more than one torque equation, with a different axis. Before diving in and writing down all the equations, think about which approach is the easiest and most direct.
There are many good examples worked out for you in the text. See pages 282-289.
Example: What is the smallest angle a ladder can make so that it does not slide?
FW
mg N
fs
Solution: We will use the condition for rotational equilibrium
0
We can choose any axis about which to take torques. The axis I choose is where the ladder touches the floor. The lever arms for the normal force and the frictional force will be zero and their torques will also be zero. Recall that the torque is
Fr
If the ladder has length L, the lever arm for the weight is the short horizontal line below the floor in the diagram. The lever arm is "the perpendicular distance from the line of the force to the point of rotation". Here it is
r
L 2
cos
The lever arm for the force of the wall pushing against the ladder is
r Lsin
Lesson 11, page 2
Lesson 11: Equilibrium, Newton's second law, Rolling, Angular Momentum (Sections 8.38.9)
Using the condition for rotational equilibrium
0
F mg 0
FW L sin
mg
L 2
cos
0
The conditions for translational equilibrium are
Fx 0 and Fy 0
Referring to the FBD, the x-components give
Fx 0
FW fs 0 FW fs
Again looking at the FBD, the y-components
Fy 0
N mg 0 N mg
Oh, no! Four equations:
Use the torque relation
FW L sin
mg
L 2
cos
0
FW fs
N mg
fs sN
FW L sin
mg
L cos 2
0
FW L sin
mg
L cos 2
FW
sin
mg 2
cos
Lesson 11, page 3
Lesson 11: Equilibrium, Newton's second law, Rolling, Angular Momentum (Sections 8.38.9)
tan mg 2FW
Substitute the two force equations from Newton's second law
tan mg 2FW
N 2 fs
fs
N 2 tan
Since this is a static friction force,
fs sN
N 2 tan
s N
1 2s tan
tan 1 2
If the coefficient of static friction is 0.4, the smallest angle is 51o.
Equilibrium in the Human Body Forces act on the structures in the body.
Example 8.10: The deltoid muscle exerts Fm on the humerus as shown. The force does two things. The vertical component supports the weight of the arm and the horizontal component stabilizes the joint by pulling the humerus in against the shoulder.
There are three forces acting on the arm: its weight (Fg), the force due to the deltoid muscle (Fm) and the force of the shoulder joint (Fs) constraining the motion of the arm.
Lesson 11, page 4
Lesson 11: Equilibrium, Newton's second law, Rolling, Angular Momentum (Sections 8.38.9)
Since the arm is in equilibrium, we use the equilibrium conditions. To use the torque equation we use a convenient rotation axis. We choose the shoulder joint as the rotation axis as that will eliminate Fs from consideration. (Why?)
0
g m 0
Fg rg Fmrm 0
Fgrg Fm sin15rm 0
Fm
Fg rg rm sin15
(30 N)(0.275m) (0.12 m)sin15
266 N
To support the 30 N arm a 270 N force is required. Highly inefficient!!
The Iron Cross. Here is an interesting video:
Because of the symmetry, half of the gymnast's weight is supported by each ring. Consider the FBD above.
0
w m 0
1 2
Wrw
Fm rm
0
Lesson 11, page 5
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