Answered: A 10.0 m long ladder weighing 50.0 N…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter10: Rotational Motion

Section: Chapter Questions

Problem 32P

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Question

a. The ladder is in equilibrium. This means that the torque around ANY pivot point must be zero. In problems like this, you can pick the pivot point of your choice, and still solve for the correct forces. However, there is typically a choice for the pivot point that makes the problem solving steps possible or at least easier. This will depend on the problem, but often depends on the fact that the torque due to a force at the pivot point is zero. This lets you ignore those torque terms in your calculations.

What is the best choice of pivot point for this problem?

Select one:

A.The base of the ladder

B.The center of mass of the ladder

C.The top of the ladder

b. By selecting the base of the ladder, you can ignore the torque due to both the normal force from the ground on the ladder and the the static friction force from the ground on the ladder.

The weight of the ladder will cause a torque about the base of the ladder. What is the magnitude of this torque? (Give your answer in N m.)

Recall: ?=??????τ=rFsinϕ

Where ?ϕ is the angle between the r vector and the force vector.

c. What is the normal force from the wall on the ladder? (Give your answer in N.)

d. Next, consider the net force in the x direction. What is magnitude of the static friction force from the ground on the ladder? (Give your answer in N.)

e. Finally, we can find the coefficient of static friction. What is this coefficient?

A 10.0 m long ladder weighing 50.0 N rests against a smooth vertical wall. If the ladder is
just on the verge of slipping when it makes a 50.0° angle with the ground find the coefficient
of static friction between the ladder and ground.
10 m
50°

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