Question

friction
page 3 of 9

μ = f_f / f_n
p171

determine the coefficient of friction for each situation. t/i
(a) it takes a horizontal force of 62 n to get a 22 kg box to just start moving across the floor. ans: 0.29
(b) it only takes 58 n of horizontal force to move the same box at a constant velocity. ans: 0.27

Explanation:

Step1: Analyze part (a)

To find the coefficient of static friction (\(\mu_s\)), we use the formula \(\mu = \frac{F_f}{F_N}\). When the box just starts moving, the friction force \(F_f\) equals the applied horizontal force, so \(F_f = 62\space N\). The normal force \(F_N\) on a horizontal surface equals the weight of the box, \(F_N = mg\), where \(m = 22\space kg\) and \(g = 9.8\space m/s^2\). Calculate \(F_N\):
\(F_N = 22\space kg \times 9.8\space m/s^2 = 215.6\space N\).

Now apply the friction formula:
\(\mu_s = \frac{F_f}{F_N} = \frac{62\space N}{215.6\space N} \approx 0.29\).

Step2: Analyze part (b)

For the coefficient of kinetic friction (\(\mu_k\)), the friction force \(F_f\) equals the applied force when moving at constant velocity (since net force is zero), so \(F_f = 58\space N\). The normal force \(F_N\) is still \(215.6\space N\) (same box, same surface).

Apply the formula:
\(\mu_k = \frac{F_f}{F_N} = \frac{58\space N}{215.6\space N} \approx 0.27\).

Answer:

(a) The coefficient of static friction is approximately \(0.29\).
(b) The coefficient of kinetic friction is approximately \(0.27\).