Question

A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 158 above the horizontal. (a) if the coefficient of static friction is 0.50, what minimum force magnitude is required from the rope to start the crate moving? (b) if mk 0.35, what is the magnitude of the initial acceleration of the crate?

Answer 1

First, illustrate the problem through a force body diagram a shown in the figure. N denotes for normal force, F for the force required to move the crate, W denotes as the weight (Wx and Wy are its vector components), and f is the frictional force which acts opposite to the direction of the crate. The supplemental angle is 22. When you use the trigo identities like sin or cos, you can either use 22 or 158. The answer would be the same.

a.) Applying Newton's 2nd law: F= ma

Summation of forces along the y-direction:
F = 0 (objects at rest)
N + Wy = 0
N + 68(9.81)(sin 22) = 0
N = 250 Newtons

Summation of forces along the x-direction
F = 0
f + F + 68(9.81)(cos 22) = 0

The frictional force's formula is f = ∪s*N (if at rest) = ∪k*N (if moving)
where ∪ is the coefficient of friction (s for static and k for kinetic)

Thus, the equation becomes

0.5(N) + F + 68(9.81)(cos 22) = 0
0.5(250) + F + 68(9.81)(cos 22) = 0
F = 743.45 Newtons

b.) Same procedure but only differs in f = ∪k*N, where ∪k = 0.35. Then,

0.35(N) + F + 68(9.81)(cos 22) = ma (for objects moving)
0.35(250) + 743.45 + 68(9.81)(cos 22) = 68(a)
a = 21.32 m/s^2