1. Static Equilibrium
AP Physics C

2. Static Equilibrium
For a system of particles to be in static equilibrium there are two conditions: it must be in translational equilibrium and it must be in rotational equilibrium.
Translational equilibrium means the system is not changing its location. For a rigid body (i.e., a solid as opposed to a liquid or gas), translational equilibrium means that at least one point in the body is stationary.
Rotational equilibrium means the system is not rotating about any axis.

3. Static Equilibrium
To be in translational equilibrium, the sum of the forces on the body must be zero, just as for a single particle. To be in rotational equilibrium, the turning effect of all the forces must also be zero. We call the turning effect of a force a torque (𝜏). For a rigid body to be in static equilibrium we must have both:
𝐹 =0 and 𝜏=0

4. Torque The door is free to rotate about an axis through O
There are three factors that determine the effectiveness of the force in opening the door:
The magnitude of the force
The position of the application of the force
The angle at which the force is applied (direction of the force)

5. 𝜏 = 𝑟 × 𝐹 (cross product of two vectors)
Torque
𝜏 = 𝑟 × 𝐹 (cross product of two vectors)
t is the torque
F is the force
symbol is the Greek tau
r is the length of the position vector
Runs from the point where torque is to be evaluated and the point of application of the force
SI unit is N.m

6. Direction of Torque Torque is a vector quantity
The direction is perpendicular to the plane determined by the position vector and the force
If the turning tendency of the force is counterclockwise, the torque will be positive
If the turning tendency is clockwise, the torque will be negative

7. Right Hand Rule
Point the fingers in the direction of the position vector
Curl the fingers toward the force vector
The thumb points in the direction of the torque

8. Multiple Torques
When two or more torques are acting on an object, the torques are added
As vectors
If the net torque is zero, the object’s rate of rotation doesn’t change

9. Check Point
The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20cm). All five horizontal forces on the stick have the same magnitude. Rank those forces according to the magnitude of the torque that they produce, greatest first.
F1: τ1 = r1 F1 sin φ1 = (20)Fsin(90o) = 20F (CCW)
F2: τ2 = r2 F2 sin φ2 = (0)Fsin(60o) = 0 (no direction)
F3: τ3 = r3 F3 sin φ3 = (20)Fsin(90o) = 20F (CW)
F4: τ4 = r4 F4 sin φ4 = (20)Fsin(60o) = 17.3F (CW)
F5: τ5 = r5 F5 sin φ5 = (80)Fsin(0o) = 0 (no direction)

10. Equilibrium Example
The woman, mass m, sits on the left end of the see-saw
The man, mass M, sits where the see-saw will be balanced
Solve for the unknown distance, x

11. Equilibrium Example
What is the normal force? m = 55 kg, M = 75 kg and the mass of the plank is 12 kg

12. Axis of Rotation
If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque
The location of the axis of rotation is completely arbitrary
Often the nature of the problem will suggest a convenient location for the axis
When solving a problem, you must specify an axis of rotation
Once you have chosen an axis, you must maintain that choice consistently throughout the problem

13. Notes About Equilibrium
A zero net torque does not mean the absence of rotational motion
An object that rotates at uniform angular velocity can be under the influence of a zero net torque
This is analogous to the translational situation where a zero net force does not mean the object is not in motion

14. Solving Equilibrium Problems
Draw a diagram of the system
Include coordinates and choose a rotation axis
Isolate the object being analyzed and draw a free body diagram showing all the external forces acting on the object
For systems containing more than one object, draw a separate free body diagram for each object

15. Problem Solving, cont. Apply the Second Condition of Equilibrium
This will yield a single equation, often with one unknown which can be solved immediately
Apply the First Condition of Equilibrium
This will give you two more equations
Solve the resulting simultaneous equations for all of the unknowns
Solving by substitution is generally easiest

16. Example of a Free Body Diagram (Forearm)
Isolate the object to be analyzed
Draw the free body diagram for that object
Include all the external forces acting on the object

17. Example of a Free Body Diagram (Beam)
The free body diagram includes the directions of the forces
The weights act through the centers of gravity of their objects

18. Example
A uniform horizontal beam with a length of l = 8.00 m and a weight of Wb = 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of  = 53 with the beam. A person of weight Wp = 600 N stands a distance d = 2.00 m from the wall. Find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam.

19. Example, cont

20. Example, cont

21. Example of a Free Body Diagram (Ladder)
The free body diagram shows the normal force and the force of static friction acting on the ladder at the ground
The last diagram shows the lever arms for the forces

22. Example mg
A uniform ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is s = Find the minimum angle  at which the ladder does not slip.

23. Example, cont.

24. Check Point
The figure gives an overhead view of a uniform rod in static equilibrium. (a) Can you find the magnitudes of unknown forces and by balancing the forces? (b) If you wish to find the magnitude of force by using a balance of torques equation, where should you place a rotation axis to eliminate from the equation? (c) The magnitude of turns out to be 65 N. What then is the magnitude of F ?

25. Sample Problem
A uniform beam, of length L and mass m 1.8 kg, is at rest on two scales. A uniform block, with mass M 2.7 kg, is at rest on the beam, with its center a distance L/4 from the beam’s left end. What do the scales read?

26. Answer

27. Example
Figure 12-6a shows a safe (mass M 430 kg) hanging by a rope (negligible mass) from a boom (a 1.9 m and b 2.5 m) that consists of a uniform hinged beam (m 85 kg) and horizontal cable (negligible mass). (a) What is the tension Tc in the cable? In other words, what is the magnitude of the force on the beam from the cable?

28. Answer