A crate of drinks of mass 20kg is place on a plane inclined at 30degree to the horizontal .if the crate slides down with a constant speed. 1, calculate the coefficient of kinematics friction.2, magnitude of friction force acting on the crate (g=10 persconds square)
weight force down slope = m g sin 30 = 20 * 9.81 * 0.5 = 98.1 Newtons
Normal force of weight on slope = m g cos 30 = 20 * 9.81 * 0.866 = 170 N
friction force up slope = 170 mu
speed constant mean force up = force down
170 mu = 98.1
we already know part B is 98.1 N
TNX bro😀😀😀😀
this is not helpful
0.866??? Is it a constant? If so what is the name of the constant? If not, how did you get that number?
Thanks in advance
To calculate the coefficient of kinematic friction (μ), we can use the following formula:
μ = tan(θ)
where θ is the angle of inclination of the plane, which in this case is 30 degrees.
1. Calculate the coefficient of kinematic friction:
μ = tan(30 degrees) ≈ 0.577
So, the coefficient of kinematic friction is approximately 0.577.
To calculate the magnitude of the friction force (F_f) acting on the crate, we can use the following formula:
F_f = μ * N
where N is the normal force acting on the crate. The normal force is equal to the weight of the crate in this case, which can be calculated using the formula:
N = m * g
where m is the mass of the crate (20 kg) and g is the acceleration due to gravity (10 m/s^2).
2. Calculate the magnitude of the friction force:
N = 20 kg * 10 m/s^2 = 200 N
F_f = 0.577 * 200 N ≈ 115.4 N
So, the magnitude of the friction force acting on the crate is approximately 115.4 N.