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A crate on a ramp will slip if the inclination is 55 degrees.

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2 Answers

Hello again, Sierra -- the steep 55 deg incline is again nearly the 3-4-5 /53 right tri ...
the weight will distribute with ~80% downhill and ~60% normal to the incline ...
static friction must counter the 80% weight downhill = Ks * 60% of weight ...
 
a) Ks is ~80/60 or with a 2/37 (5%) adjust -> Ks ~84/57 ~1 27/57 ~1 9/19 ~1.45
 
b) friction counters the 80% downhill => 5kg weighs 50N ... 80% is 40N ... Regards :)
(a) Let's call the direction parallel to the incline the x-direction and the direction perpendicular to the incline the y-direction. The total force in either direction is zero when the crate starts to slip:
 
∑Fx= 0,    ∑Fy= 0
 
The forces perpendicular to the incline are the component of gravity, mg cos(55) (down), and the normal force, N (up), so that
∑Fy= 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, Fs, (up):
 
∑Fx= Fs- mg sin(55) =0
 
 
For static friction, you use the maximum value because the crate is about to slip:
 
Fs_max=µ N =µ mg cos(55),
 
where µ is the coefficient of friction we are looking for. Substitute this in and get
 
 
∑Fx= 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 Fs for m=5 kg, substitute this and µ=1.43 into
 
Fs=µ mg cos(55) = 1.43*5*9.8 cos(55) = 40.1 newtons up the incline