Question

1. a 38 kg crate rests on a horizontal floor. a horizontal pulling force of 170 n is needed to start the crate moving. after it is in motion, a horizontal pulling force of 120 n is needed to slide the crate across the floor at a constant speed.

a. find the coefficient of static friction between the crate and the floor
b. find the coefficient of kinetic friction between the crate and the floor

Explanation:

Step1: Calculate normal force

The normal force \(N\) on the crate is equal to its weight. Using \(N = mg\), where \(m = 38\space kg\) and \(g=9.8\space m/s^{2}\).
\(N=38\times9.8 = 372.4\space N\)

Step2: Find coefficient of static friction (\(\mu_s\))

The maximum static - friction force \(f_{s,max}\) is equal to the force required to start the motion. The formula for static - friction force is \(f_{s,max}=\mu_sN\). Given \(f_{s,max} = 170\space N\) and \(N = 372.4\space N\).
\(\mu_s=\frac{f_{s,max}}{N}=\frac{170}{372.4}\approx0.46\)

Step3: Find coefficient of kinetic friction (\(\mu_k\))

When the crate is in motion at a constant speed, the kinetic - friction force \(f_k\) is equal to the pulling force. The formula for kinetic - friction force is \(f_k=\mu_kN\). Given \(f_k = 120\space N\) and \(N = 372.4\space N\).
\(\mu_k=\frac{f_k}{N}=\frac{120}{372.4}\approx0.32\)

Answer:

a. The coefficient of static friction \(\mu_s\approx0.46\)
b. The coefficient of kinetic friction \(\mu_k\approx0.32\)