(a) Let's call the direction parallel to the incline the xdirection and the direction perpendicular to the incline the ydirection. The total force in either direction is zero when the crate starts to slip:
∑F_{x}= 0, ∑F_{y}= 0
The forces perpendicular to the incline are the component of gravity, mg cos(55) (down), and the normal force, N (up), so that
∑F_{y}= N mg cos(55) =0,
or N=mg cos(55)
The forces parallel to the incline are the component of gravity, mg sin(55) (down) and the force of static friction, F_{s}, (up):
∑F_{x}= F_{s} mg sin(55) =0
For static friction, you use the maximum value because the crate is about to slip:
F_{s_max}=µ N =µ mg cos(55),
where µ is the coefficient of friction we are looking for. Substitute this in and get
∑F_{x}= µ mg cos(55)  mg sin(55) =0
Cancel mg and solve for µ:
µ cos(55)  sin(55) =0
µ= sin(55)/cos(55) =tan(55) ≈1.43
Note that µ is independent of the mass!
(b) To find F_{s} for m=5 kg, substitute this and µ=1.43 into
F_{s}=µ mg cos(55) = 1.43*5*9.8 cos(55) = 40.1 newtons up the incline
10/16/2013

Andre W.