Question

g A dockworker loading crates on a ship finds that a 22-kg crate, initially at rest on a horizontal surface, requires a 72-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 52 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

Answer 1

Answer:

μs = 0.333 and μk = 0.24

Explanation:

Mass of crates = 22kg

F = 72newton

F =μR

Also given that g = 9.8m/s^2 which is that acceleration due to gravity

Hence R = N = mg

= 9.8 × 22

= 215.6 newton

F is frictional force

μ is coefficient of friction

N or R is normal reaction

Hence, the static friction

F =μsR

72 = μs × 215.6

μs = 72/215.6

μs = 0.333

The kinetic friction coefficient

F =μkR

When F = 52N and R = N = 215.6N

52 = μk × 215.6

μk = 52/215.6

μk = 0.24

Answer 2

Answer:

μs = 0.333

μk = 0.241

Explanation:

a) The coefficient of static friction is calculated by using the following formula:

$F=\mu_sN$    (1)

μs: coefficient of static friction

N: normal force

F: force exerted on the box where the box is at rest = 72N

The normal force is given by N=Mg with M the mass of the box (22kg) and g the gravitational acceleration (9.8m/s^2). You replace these values in the equation (1) and solve for μs:

$72N=\mu_s(22kg)(9.8m/s^2)=\mu_s(215.6N)\\\\\mu_s=\frac{72N}{215.6N}\\\\\mu_s=0.333$

The coefficient of static friction is 0.333

b) The coefficient of kinetic friction is calculated by using the same equation (1), but with the value for F needed to maintain the box in motion.

$F=\mu_kN=\mu_kMg$     (2)

F: force exerted on the box when it is already in motion

μk: coefficient of kinetic friction

You solve the equation (2) for μk:

$\mu_k=\frac{F}{Mg}=\frac{52N}{215.6N}=0.241$

The coefficient of kinetic friction is 0.241