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

a.) What is the coefficient of static friction?
b.) The block has a mass of 5 kg. What is the force of friction on it before it begins to slip?

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Brad M. | Summer Online Finance Specialist: WACC NPV DCF TVM YTMSummer Online Finance Specialist: WACC ...
4.9 4.9 (233 lesson ratings) (233)
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 :)
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
(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