a crate rest on a ramp that makes an angle of tangent with the horizontal.the coefficient of kinetic friction is 0.35 and the coefficient of static friction is 0.45.the inclination of the ramp is slowly increased.(a)find the angle at which the crate starts to slip down the ramp.(B)find the angle at which the crate slips down at constant velocity once motion has started.
To find the answer to this question, we need to understand the concept of friction and equilibrium.
(a) To determine the angle at which the crate starts to slip down the ramp, we need to compare the gravitational force pulling the crate downhill with the maximum static friction force opposing the motion.
First, let's find the maximum static friction force:
Static friction force (F_static) = coefficient of static friction (μ_static) * normal force (N)
The normal force (N) acting on the crate is the component of the crate's weight perpendicular to the ramp's surface. The normal force can be calculated as:
N = weight of the crate * cos(θ)
where θ is the angle of the ramp with respect to the horizontal.
The weight of the crate can be determined by multiplying the mass of the crate (m) by the acceleration due to gravity (g):
Weight of the crate = m * g
Now, let's substitute these values into the equation for the maximum static friction force:
F_static = μ_static * m * g * cos(θ)
To determine the angle at which the crate starts to slip, the maximum static friction force should be equal to the gravitational force acting downhill. The gravitational force (F_gravity) can be calculated as:
F_gravity = m * g * sin(θ)
Equating these forces, we have:
μ_static * m * g * cos(θ) = m * g * sin(θ)
Dividing both sides by m * g, we get:
μ_static * cos(θ) = sin(θ)
Now, we can solve for θ:
μ_static = tan(θ)
Take the inverse tangent of both sides:
θ = arctan(μ_static)
Substituting the given coefficient of static friction:
θ = arctan(0.45)
Using a calculator, we find:
θ ≈ 24.28 degrees
Therefore, the angle at which the crate starts to slip down the ramp is approximately 24.28 degrees.
(b) To find the angle at which the crate slips down at constant velocity once the motion has started, we need to consider the kinetic friction force.
The kinetic friction force (F_kinetic) can be calculated as:
F_kinetic = coefficient of kinetic friction (μ_kinetic) * normal force (N)
The normal force (N) is the same as calculated before.
Using the same concepts as before, we equate the gravitational force acting downhill with the kinetic friction force:
μ_kinetic * m * g * cos(θ) = m * g * sin(θ)
Dividing both sides by m * g:
μ_kinetic * cos(θ) = sin(θ)
Now, solve for θ:
μ_kinetic = tan(θ)
Take the inverse tangent of both sides:
θ = arctan(μ_kinetic)
Substituting the given coefficient of kinetic friction:
θ = arctan(0.35)
Using a calculator, we find:
θ ≈ 19.29 degrees
Therefore, the angle at which the crate slips down at constant velocity once the motion has started is approximately 19.29 degrees.