A 15kg box is resting on a hill forming an angle with the horizontal. The coefficient of static friction for the box on the surface is 0.45. Calculate the maximum angle of the incline before the box starts to move.

Question

Steps included would be great, I have the answer in my textbook but I would like to know how to do it as I have an exam tomorrow. Thanks.

Answer 1

let Θ be the angle
... n is the normal force of the box against the hill
... μ is the coefficient of static friction

n = m g cos(Θ)

frictional force parallel to the slope = μ n = μ m g cos(Θ)

gravitational force parallel to the slope = m g sin(Θ)

when the gravitational force becomes greater, the box moves

m g sin(Θ) > μ m g cos(Θ) ... tan(Θ) > μ

Answer 2

call the angle A
normal force on roadway = m g cos T
so
max friction force = 0.45 * m g cos T

component of weight down roadway = m g sin T

when the friction force = m g sin T, wooosh

0.45 m g cos T = m g sin T

or sin T/cos T = tan T = 0.45

Answer 3

Ah, that angle I meant to call A I actually called T

Answer 4

M*g = 15 * 9.8 = 147 N. = Wt. of box,
Fp = 147*sin A = Force parallel to the hill,
Fn = 147*Cos A = Normal force,
Fs = u * Fn = 0.45 *147*CosA = 66.2*CosA = Force of static friction,

Fp-Fs = M*a,
147*sin A - 66.2*CosA = 15*0,
66.2*CosA = 147*sinA,
147*sinA/CosA = 66.2, sinA/CosA = TanA,
TanA = 66.2/147 = 0.45,
A = 24.2o = Max. angle of incline.