A box rests on a rough board 10.0meters long. When one end of the board is slowly raised to a height of 6.0meters above the other end, the box begins to slide.

Question

1. What is the coefficient of static friction?
2. Once you have claculated the coefficient of static friction, do you have sufficient information to calculate the acceleration of the box? If not list the additional information you need, else determine the acceleration.

Answer 1

first geometry
3,4,5 triangle
sin A = 0.6 so A = 36.9 degrees slope
cos A = 0.8
now problem
normal force = m g cos A = .8 m g
friction force = mu m g cos A = mu(.8m g)
force down slope = m g sin A = .6 m g
begins to slide so forces equal and opposite
.8 mu m g = . 6 m g
mu = 3/4 = .75

Well, the net force is zero so it does not accelerate BUT if it gets shook a little off it goes because kinetic coef of friction is less than static. In other words this situation is unstable. To get the acceleration we need the kinetic coef of friction.

Answer 2

1. Mg = Wt. of box.

Fp = Mg*sin A = Mg*(6/10) = 0.6Mg = Force parallel to ramp.

X^2 + 6^2 = 10^2, X = 8.

Fn = Mg*CosA = Mg*(8/10) = 0.8Mg = Normal force.

Fs = u*Fn = u*0.8Mg = Force of static friction.

Fp-Fs = M*a.
0.6Mg-u*0.8Mg = M*0 = 0,
Divide both sides by Mg:
0.6 - 0.8u = 0, u = 0.75.

Answer 3

2. Fp-Fk = M*a.
0.6Mg - u*0.8Mg = M*a,
Divide both sides by M:
0.6g - ug = a,
0.6*9.8 - 9.8u = a,
-9.8u = a - 5.88,
u = -0.1a + 0.6.

We have 1 Eq and 2 unknowns. Therefore, we cannot calculate "a".

Answer 4

1. hyp(AC) = 10m, alt(AB)= 6m, base(BC) = 10^2 - 6^2 = 8m
Let C be the angle theta <ABC = 90 degree
coefficient of static friction = tan(C)
so, find sin(C) and cos(C)
sin(C) = 0.6
cos(C) = 0.8
tan(C) = sin(C)/cos(C) = 0.6/0.8 = 0.75