Asked by Anonymous
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)
Answers
Answered by
Anonymous
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
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
Answered by
praise
TNX broππππ
Answered by
Anonymous
this is not helpful
Answered by
Case
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
Thanks in advance
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