A 32 kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to 23.2° the crate begins to slide downward.

(a) What is the coefficient of static friction between the crate and the ramp?

(b) At what angle does the crate begin to slide if its mass is doubled?

I have no idea how to find part a.

For part B why is it not twice the angle?

component of the weight down the ramp = m g sin 23.2

component of the weight perpendicular to the ramp = m g cos 23.2 (this is the normal force)
friction force = coef * normal force = weight component down ramp when slip starts
so
mu m g cos 23.2 = m g sin 23.2
solve for mu, the coefficient
Notice that the mass cancels out, it does not matter.

To find the coefficient of static friction in part (a), we can use the concept of equilibrium. The crate is at rest on the inclined ramp, which means the gravitational force pulling the crate downwards is balanced by the static friction force acting on it in the opposite direction along the ramp. The formula for the static friction force is:

fs = μs * N

Where fs is the static friction force, μs is the coefficient of static friction, and N is the normal force exerted by the ramp on the crate. For a crate on an inclined plane, the normal force is given by:

N = mg * cos(θ)

Where m is the mass of the crate, g is the acceleration due to gravity (approximately 9.8 m/s^2), and θ is the angle of the ramp with the horizontal.

Now, to find the coefficient of static friction, we need to set up an equation for equilibrium in the perpendicular direction to the ramp. The equation is:

N = mg * cos(θ)

Using this equation, we can substitute the expression for N in terms of m, g, and θ into the equation for the static friction force:

fs = μs * N = μs * mg * cos(θ)

Now, we can solve for the coefficient of static friction (μs) by isolating it:

μs = fs / (mg * cos(θ))

To find the value of μs, we need to know the angle at which the crate begins to slide. This information is not provided in the question.

Moving on to part (b), when the mass of the crate is doubled, the force of gravity pulling the crate down the ramp will also double (since F = mg). As a result, the friction force needed to balance it must also double. However, the maximum static friction force is directly proportional to the normal force, which depends on the angle θ. Therefore, when the mass is doubled, the angle at which the crate begins to slide will not be twice the original angle. The actual relationship between the angles can be determined by setting up and solving equations using the concepts described above.