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.
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.
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
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
Ah, that angle I meant to call A I actually called T
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.
To find the maximum angle of the incline before the box starts to move, we need to consider the forces acting on the box.
First, we have the force due to gravity pulling the box downward, which can be calculated as Fg = m * g, where m is the mass of the box (15 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Next, we have the normal force (Fn) acting perpendicular to the surface of the incline. This force is equal in magnitude and opposite in direction to the force due to gravity, so Fn = Fg = m * g.
The force of static friction (Fs) opposes the motion of the box and prevents it from sliding down the incline. The maximum static friction force is given by the equation Fs(max) = coefficient of static friction * Fn.
Since Fn = Fg, we can substitute this into the equation to get Fs(max) = μ * m * g, where μ is the coefficient of static friction (0.45).
Now, to determine the maximum angle of the incline, we can set Fs(max) equal to the component of the force due to gravity acting parallel to the incline. This component can be calculated as Fp = m * g * sinθ, where θ is the angle of the incline.
Therefore, we have μ * m * g = m * g * sinθ.
Simplifying the equation, we find μ = sinθ.
To solve for θ, we need to take the inverse sine (sin^(-1)) of both sides of the equation:
sin^(-1)(μ) = θ.
Using the given coefficient of static friction μ = 0.45, we can substitute this value to find:
θ = sin^(-1)(0.45).
Using a calculator, the maximum angle of the incline is approximately 27.02 degrees.
Therefore, the maximum angle before the box starts to move is approximately 27.02 degrees.