Question

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)

Answer 1

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

Answer 2

TNX bro😀😀😀😀

Answer 3

this is not helpful

Answer 4

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