Question

An 80 N box initially at rest is pulled by a horizontal rope on a horizontal table. The coefficients of kinetic and static friction between the box and the table are 14 and 12, respectively. What is the friction force on this box if the tension in the rope is

a. 0N?

b.30N?

c.39N?

d.41N?

e.150N?

Answer 1

Solution:

Given: Weight of box (W) = 80 N

The coefficients of kinetic friction () = 1/4 = 0.25

The coefficients of static friction () = 1/2 = 0.5

So, Normal force (N) = weight (W) = 80 N

friction is equal to the pull, unless the pull is greater than the static friction,

then it becomes: f_s = N = (0.5) (80 N) = 40 N

So the first three parts frictional force will be the same as the pull or:
(a) 0 N
(b) 30 N
(c) 40 N

for d and e, it will be : f_k = N = (0.25) (80 N) = 20 N
(d) 20 N
(e) 20 N