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

**1.45**

**40N**... Regards :)

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?

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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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:

∑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

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