Question

A block is at rest on an inclined plane whose elevation can be varied. The coefficient of static friction is μ_s= 0.45, and the coefficient of kinetic friction is μ_k = 0.18. The angle of elevation θ is increased slowly from the horizontal.

At what value of θ does the block begin to slide (in degrees)?

What is the acceleration of the block?

Answer 1

At what value of θ does the block begin to slide (in degrees)?

The coefficient of static friction only plays a role.

The force of gravity along the inclined plane at angle θ is

F_grav = m g sin(θ)

The force of static friction equals

F_fric = μs F_normal = μs m g cos(θ)

The block will start sliding just after these are equal. The angle at which this happens can thus be obtained as

m g sin(θ) = μs m g cos(θ)

sin(θ)/cos(θ) = μs

tan(θ) = μs

θ = arctan(μs)

= arctan(0.45) = 24.2 degrees

What is the acceleration of the block?

Once the block slides, the friction is determined by μk.

The net force is a sum of two components, the force of gravity along the incline and the friction along the incline but opposite. This sum must equal mass time acceleration (Newton 2):

m g sin(θ) - μk m g cos(θ) = m a

So

a = g sin(θ) - μk g cos(θ)

= 9.81 m/s^2 * sin(24.2) - 0.18 * 9.81 m/s^2* cos(24.2)

= 9.81 m/s^2 * 0.41 - 0.18 * 9.81 m/s^2* 0.91

= 4.02 – 1.61 = 2.41 m/s^2