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Can you predict the angle at which an inclined plane can be inclined if an object is placed on it and all you know is the coefficient of friction between the object and the surface of the plane?
Wiki User
∙ 16y ago
The frictional force that opposes the motion of an object on an inclined plane is given by the formula:
Ffriction = (mu)N,
where mu (the Greek letter mu) is the coefficient of friction, and N is the Normal force, which is the force equal and opposite to the component of the object's weight perpendicular to the surface of the incline.
The Normal force will be equal to Wcos(theta), where W is the weight of the object (W= mg) and theta is the angle of the incline.
When motion down the plane is impending (that is, a split second before the friction is overcome and the object starts to slide down the plane), Ffriction is equal and opposite to the component of the weight parallel to the surface of the plane. That component is equal to Wsin(theta).
So, what does that give us?
We know that
(1) Ffriction = Wsin(theta)
(2) Ffriction = (mu)N
(3) N = Wcos(theta)
Substituting for N in equation (2) gives us
Ffriction = (mu)Wcos(theta).
Equating equ. (1) and (2) gives us
muWcos(theta) = Wsin(theta).
Solving for mu gives us
mu = sin(theta)/cos(theta)
mu = tan(theta)
theta = tan-1(mu) or theta = arctan(mu)
So, the arctangent of mu is the angle of incline.
(I guess I coulda just said that right from the beginning.)
Wiki User
∙ 16y ago
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I have no clue.
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Predict the effect that significant friction would have on the acceleration of the box as it slides down the incline. Explain the cause of the predicted effect?
I have no clue.
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