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Chapter 4: Problem 95

A box full of books rests on a wooden floor. The normal force the floor exerts on the box is \(250 \mathrm{N}\). (a) You push horizontally on the box with a force of \(120 \mathrm{N},\) but it refuses to budge. What can you say about the coefficient of static friction between the box and the floor? (b) If you must push horizontally on the box with a force of at least \(150 \mathrm{N}\) to start it sliding, what is the coefficient of static friction? (c) Once the box is sliding, you only have to push with a force of \(120 \mathrm{N}\) to keep it sliding. What is the coefficient of kinetic friction?

Short Answer

Expert verified
Answer: The coefficient of static friction is 0.6, and the coefficient of kinetic friction is 0.48.

Step by step solution

01

Determine the maximum static friction

To determine the maximum static friction between the box and the floor, we need to know the relationship between the friction force and the normal force. The maximum static friction force \(f_s\) can be expressed as: $$ f_s = \mu_s \times N $$ where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force. In this case, the normal force is \(250 \mathrm{N}\).
02

Solve for the coefficient of static friction (a)

Since the box refuses to budge with a \(120 \mathrm{N}\) horizontal force, we know that the maximum static friction force is at least equal to this value, so we can write the inequality: $$ \mu_s \times 250 \mathrm{N} \ge 120 \mathrm{N} $$ Solving for the coefficient of static friction, we have: $$ \mu_s \ge \frac{120 \mathrm{N}}{250 \mathrm{N}} = 0.48 $$ So, the coefficient of static friction must be greater than or equal to \(0.48\).
03

Solve for the coefficient of static friction (b)

The problem states that you must push horizontally with a force of at least \(150 \mathrm{N}\) to start the box sliding. This gives us the equation: $$ \mu_s \times 250 \mathrm{N} = 150 \mathrm{N} $$ Solving for the coefficient of static friction, we get: $$ \mu_s = \frac{150 \mathrm{N}}{250 \mathrm{N}} = 0.6 $$ Therefore, the coefficient of static friction is \(0.6\).
04

Solve for the coefficient of kinetic friction (c)

Once the box is sliding, you only have to push with a force of \(120 \mathrm{N}\) to keep it sliding. This means that the kinetic friction force \(f_k\) is equal to the applied force: $$ f_k = \mu_k \times N $$ Since the applied force is equal to the friction force, we can set them equal to each other: $$ 120 \mathrm{N} = \mu_k \times 250 \mathrm{N} $$ Solving for the coefficient of kinetic friction, we get: $$ \mu_k = \frac{120 \mathrm{N}}{250 \mathrm{N}} = 0.48 $$ Therefore, the coefficient of kinetic friction is \(0.48\).

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