1. Chapter 9
Torque

2. Student learning objectives
To extend the particle model to the extended object (rigid-body) model.
To understand the static equilibrium of an extended rigid body.

3. Opening a door
Consider the common experience of pushing open a door. Shown is a top view of a door hinged on the left. Four pushing forces are shown, all of equal strength. Which of these will be most effective at opening the door?
The ability of a force to cause a rotation depends on three factors:
the magnitude F of the force.
the radial distance r from the point of application of the force to the hinge, or pivot.
the angle at which the force is applied.

4. Some vocabulary
Line of action, line of force (F) (blue dashed line) – the line along which the force acts.
Axis of rotation (hinge, pivot) – in the examples above, the axis of rotation is out of the page (z axis).

5. More Vocabulary
Radial axis (r), all or part of the top of the door - the radial axis is the distance between the hinge and the application of the force. The radial axis due to F4 is less than that of the other forces.

6. More Vocabulary
Lever arm (ℓ) (moment arm) – red dashed line – the line that makes a 900 between the axis of rotation and the line of action. In the first diagram, the lever arm and the radial axis coincide. In the third diagram, there is no lever arm.

7. Magnitude of Torque = (Magnitude of the force) x (Lever arm)
ℓ = r sin θ
Where θ is the angle between the line of action and the radial axis
Direction of Torque: The torque is positive when the force tends to produce a counterclockwise rotation about the axis.
SI Unit of Torque: newton x meter (N·m) Not the same a Joule!

8. 9.1 The Action of Forces and Torques on Rigid Objects
The amount of torque depends on where and in what direction the
force is applied, as well as the location of the axis of rotation.

9. Rank in order, from largest to smallest, the five torques τa − τe
Rank in order, from largest to smallest, the five torques τa − τe. The rods all have the same length and are pivoted at the dot.
(a)
(b)
(c)
(d)
(e)
Answer: A

10. EOC #8 – Extended body
The left end of the meter stick is pinned to the table so the stick can rotate freely parallel to the table top. Two forces, both parallel to the table are applied to the stick. The net torque is zero. The forces and angles are shown. How far from the pivot point (or axis of rotation) is the 6.00 N force applied?

11. EOC #8 - Analysis
The drawing below acts as an extended free body diagram. The weight force and the normal force from the table are parallel to the axis of rotation (pivot point), and provide no torque. The 6.00 N force tends to cause cw rotation, so the torque about the pivot from that force is negative. The 4.00 N force tends to cause ccw rotation, so the torque about the pivot from that force is positive. Since the problem states the net torque is zero, these two torques must have an equal magnitude.
knowns Find r2 F1 = 4.00N θ1 = 90° F2 = 6.00 N θ2 = 60° r1 = 1 m
ז= Fr(sin θ)

12. EOC #8 - Answer Σז = 0 ז1 – ז2 = 0 F1r1(sin θ1) – F2 r2 (sin θ2) = 0
4 N(1m) – 6N (r sin 60°) = 0
r = 0.77 m
knowns Find r2 F1 = 4.00N θ1 = 90° F2 = 6.00 N θ2 = 60° r1 = 1 m
ז= Fr(sin θ)

13. EQUILIBRIUM OF A RIGID BODY
A rigid body is in equilibrium if:
This analysis will most likely involve multiple equations with more than one unknown.

14. DEFINITION OF CENTER OF GRAVITY
The center of gravity(cg) of a rigid body is the location at which its weight can be considered to act when the torque due to the weight is being calculated.
An object of uniform density (“uniform object”) has a cg at its center.

15. Reasoning Strategy
1. Select the object to which the equations for equilibrium are to be applied.
2. This is the most important step. Draw an extended free-body diagram that shows all of the external forces acting on the object. You will no longer be able to use a particle to represent the object.
3. The weight force acts at the center of gravity. Assume the center of gravity is at the mid point of the object, unless told otherwise.
4. Apply Newton’s 1st Law (since ma=0 for equilibrium) in component form.
5. Select a pivot point where one or more of the unknown forces will have a torque of zero. Set the sum of the torques about this axis equal to zero. The pivot you pick does not have to be an actual axis of rotation for this object!
6. Solve the equations for the desired unknown quantities.

16. EOC #12
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the right support (see the drawing). To what distance x can a person who weighs 380 N walk on the overhanging part of the plank before it just begins to tip?

17. EOC #27
A man holds a 178-N ball in his hand. The forearm weighs 22 N and has a cg as shown.
Find the magnitude of M
Find the magnitude and direction of the force applied by the upper arm bone to the forearm at the elbow joint

18. EOC #27
A man holds a 178-N ball in his hand. The forearm weighs 22 N and has a cg as shown.
Find the magnitude of M
Find the magnitude and direction of the force applied by the upper arm bone to the forearm at the elbow joint
M = 1.21 x 103 N
F (bone) = 1.01 x 103

19. EOC 29
An inverted “V” is made of uniform boards and weights 356 N. Each side has the same length and makes the angle shown with the vertical. Find the magnitude of the static frictional force that acts on the lower end of each leg of the “V”.

20. EOC 29
An inverted “V” is made of uniform boards and weighs 356 N. Each side has the same length and makes the angle shown with the vertical. Find the magnitude of the static frictional force that acts on the lower end of each leg of the “V”. Fb Fn Fg fs
Since ΣF = 0 in both directions:
Fn = Fg = 356N/2
Fb = fs
Find: fs

21. EOC 29 For the board to be in static equilibrium:
Σ ז about ANY axis = 0. Therefore pick an axis of rotation which:
Gets rid of as many unknowns as possible
Keeps the unknown we are looking for or something equal to it. Fb Fn Fg fs
Since ΣF = 0 in both directions:
Fn = Fg = 356N/2
Find: Fb = fs

22. EOC 29 Since ΣF = 0 in both directions: Fn = Fg = 356N/2 Find: Fb = fs
Let’s choose the point where the left leg touches the floor Σ זfloor = 0. Fn and fs provide 0 torque since they act towards the pivot point. Σ זfloor = Fb(L cos 30)-Fg(L/2 sin 30) = 0 Fb = fs = 51.4 N Fb Fn Fg fs
Since ΣF = 0 in both directions:
Fn = Fg = 356N/2
Find: Fb = fs

23. Will the ladder slip?
A 3.0-m long ladder leans against a frictionless wall at an angle of 60°. What is the minimum value of us that prevents the ladder from slipping?

24. Will the ladder slip?
A 3.0-m long ladder leans against a frictionless wall at an angle of 60°. What is the minimum value of us that prevents the ladder from slipping?

25. Will the ladder slip?
A 3.0-m long ladder leans against a frictionless wall at an angle of 60°. What is the minimum value of us that prevents the ladder from slipping?
Answer:
us-min = ½ tan 30° = 0.29