1. A ladder with length L weighing 400 N rests against a vertical frictionless wall as shown below. The center of gravity of the ladder is at the center of the ladder. Find the coefficient of static friction s between the ladder and the horizontal ground at its base if an 80 kg man climbs one-third up before the ladder starts to slip.
No Friction L y
60 x

2. Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10
Question 11

3. C : where is the the torque about any chosen point.
Problems such as the one at hand require analyzing the statics of rigid bodies. The situation described on the first slide is static as long as the man is not more than one-third of the way up the ladder, because there is no motion. Therefore, the conditions for static equilibrium apply to the situation. What are the static equilibrium conditions?
A :
B :
C : where is the the torque about any chosen point.
D : all of the above

4. Choice: A
Incorrect
This is only one of the required conditions for static equilibrium. It is true that the sum of the forces along the x-axis must be zero, but there are more conditions that must be met.

5. Choice: B
Incorrect
This is only one of the required conditions for static equilibrium. It is true that the sum of the forces along the y-axis must be zero, but there are more conditions that must be met.

6. Choice: C
Incorrect
This is only one of the required conditions for static equilibrium. The net torque on the ladder must be zero, but other conditions must apply.

7. Choice: D
Correct
All three of these conditions must be satisfied in order for a system to be in static equilibrium.

8. 2. Which one of these diagrams correctly shows all of the forces acting on the ladder?Nw
Ff = Static Friction Force
WL=Weight of ladder
Wm=Weight of man
Ng=Normal force from the ground
Nw=Normal force from the wall A: WL Wm Ff Nw WL B: Wm Ng Ff WL Wm C: Ng Ff

9. Choice: A
Incorrect
If the ladder slips, the base will move away from the wall. The force of friction between the ladder base and the ground will resist this motion. The arrow representing friction should point in the opposite direction. Also, there is a normal force from the ground pushing on the ladder.

10. Choice: B
Correct
This diagram correctly shows all of the forces involved in the situation.
NOTE: The force here called “weight of the man” is actually a normal force acting on the ladder. It is directed downwards because the man is standing on a horizontal step. The actual force “weight of the man” acts on the man, of course.

11. There is a normal force from the wall pushing on the ladder.
Choice: C
Incorrect
There is a normal force from the wall pushing on the ladder.

12. 3. Applying the first condition for static equilibrium, we find which of the following to be true for the sum of forces in the x-direction? A: B: C:

13. Choice: A
Correct
This sums up all the forces that act in the positive and negative x-directions. Notice that the static friction force applied by the ground is equal to the normal force from the wall pushing on the ladder.

14. Choice: B
Incorrect
Check the diagram from the previous question. The normal force from the ground pushing on the ladder acts in the positive y-direction.

15. Choice: C
Incorrect
Check the diagram from the previous question. The force of gravity is directed in the negative y-direction. Therefore, the weight of the ladder and the man are not included in this sum.

16. 4. Once again, let’s apply the first condition for static equilibrium, but this time for the forces in the y-direction. Which of the following is correct? A: B: C:

17. Choice: A
Incorrect
Check the diagram from the question 2. The normal force from the wall pushing on the ladder acts in the negative x-direction, so it should not be involved in this sum.

18. Choice: B
Correct
This sums up all the forces that act in the positive and negative y-directions. Notice that the normal force due to the ground pushing on the ladder is equal to the added weights of the man and the ladder.

19. Choice: C
Incorrect
The sum of the forces in the y-directions should include the weight of the man and the ladder. The normal force from the wall acting on the ladder should not be included in the sum, because it acts in the negative x-direction.

20. 5. If the ladder is on the verge of slipping, then Ff should assume its maximum value. Which of these expressions gives this value? (where is coefficient of static friction) A: B: C:

21. Choice: A
Correct
Notice that choice B is also true, because from the previous question we saw that:
Choice B is the better choice, because it is more general. Choice A is correct based on an analysis of the details of this particular problem.

22. Choice: B
Correct
The maximum static friction force is equal to the coefficient of static friction times the normal force that is perpendicular to the direction of the friction. Notice that choice A is also true, because from the previous question we saw that:
Choice B is the better choice, because it is more general. Choice A is correct based on an analysis of the details of this particular problem.

23. Choice: C
Incorrect
The normal force relevant here is the normal force exerted by the ground, because it acts in a direction perpendicular to the force of friction.

24. Notice the following:
Since in this problem we are only concerned with the ladder at the critical point before it slips.
From combing our findings from questions 3,4, and 5.

25. A: The axis of rotation and the angular velocity.
6. We now have two unknowns, Nw and s, but we only have one equation. Let’s try using the second condition for static equilibrium to find another helpful equation: Which of the following concepts/values will be important to find the net torque on the system?
A: The axis of rotation and the angular velocity.
B: Lever arm for each force and moment of inertia.
C: The axis of rotation and lever arm for each force.

26. Choice: A
Incorrect
Angular velocity is useless in this situation. Remember, the ladder is not in motion.

27. Choice: B
Incorrect
We don’t need to know the moment of inertia. Torque can be found in other situations using the relation: Where I is moment of inertia and  is angular acceleration. This form of expressing torque will not be useful here, because it does not allow us to use the variables that we are trying to find the values of.

28. Choice: C
Correct
A torque can be found by multiplying a force by it’s lever arm. We need to now choose an axis of rotation so that we know which forces and corresponding lever arms to use to calculate the net torque.

29. 7. How is the lever arm for each force defined?
A: It is the distance from the center of mass of the object under consideration to each of the forces.
B: It is the minimum distance from the axis of rotation to the point at which the force is applied.
C: It is the distance from the axis of rotation to the center of mass of the ladder.

30. Choice: A
Incorrect
Torque involves a force that gives an object the tendency to rotate about a certain point, or an axis of rotation. Only the component of a force that is perpendicular to the distance between the location where the force is applied and the axis of rotation cause this rotation. A lever arm is the minimum distance from the axis of rotation to the point at which the force is applied.
Lever arm
Perpendicular component of an applied force
Axis of rotation

31. This is the length that would contribute to a torque.
Choice: B
Correct
This is the length that would contribute to a torque.
NOTE:
A torque will cause rotation only if it is not compensated by countertorques.

32. Choice: C
Incorrect
The lever arm is the minimum distance from the axis of rotation to the point at which the force is applied. Don’t Forget: Only the component of a force that is perpendicular to the distance between the location where the force is applied and the axis of rotation cause this rotation.
Component of the applied force that causes torque and the rotation of the object.
Lever arm
Axis of rotation
Applied force

33. A: Nw B: Wm and WL C: Ng and Ff
8. We can choose to place our axis of rotation anywhere on the ladder, but it is good practice to choose an axis through which more than one force passes. So in this case, we choose our axis to be where the ladder contacts the ground. Choosing this axis would make which of the forces’ lever arm zero?
A: Nw
B: Wm and WL
C: Ng and Ff

34. Choice: A
Incorrect
The perpendicular component of this normal force has a non-zero lever arm and produces a torque about the axis.
Nwsin(60)
Front side view of ladder Nw
Axis of rotation
Axis of rotation

35. These weights produce torques about the chosen axis.
Choice: B
Incorrect
These weights produce torques about the chosen axis. _2 L _3
WLcos (60)
Axis of rotation
Wmcos (60)

36. Choice: C
Correct
These forces act at a point on the axis, so their corresponding lever arms are zero.

37. 9. Now that we have chosen the axis and noted that the magnitude of torque is equal to the force times the lever arm, which expression will we get when we apply the second condition to the system? A: B: C:

38. Choice: A
Incorrect
None of these forces are perpendicular to the ladder. Only forces that act perpendicular to the ladder or perpendicular components of the forces that act on the ladder will contribute to a net torque on the ladder. Check the hints from the previous question.

39. Choice: B
Incorrect
The lever arms associated with the weights are not correct. The components of the weights which are perpendicular to the ladder should be used.

40. Choice: C
Correct
This equation contains the correct lever arms and forces. Also, the signs of the torques are correct.

41. 10. Solve the correct equation from question 9 for the normal force from the wall pushing on the ladder Nw.
Answer

43. 11. Calculate the value of the coefficient of static friction between the ground and ladder by setting the two equations for Nw together, solving for s, inserting the given values, and simplifying.

44. Answer: Set equations for Nw equal to one another: Solve for s
Insert given values and simplify

45. Reflection problem
1. How would the problem change if the floor was frictionless?

46. Reflection problem
2. If the man was only half as heavy, would he be able to go farther up without the ladder falling down?