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Sitting on a horizontal track is a mass of 10kg. One end of the track is raised until the mass just begins to slip. If the coefficient of static friction is .4, what is the minimum angle at which the mass begins to slip?

2006-09-04 04:16:23 · 1 answers · asked by sur2124 4 in Education & Reference Homework Help

1 answers

If an object is at rest on a surface and you push against it and it doesn't move, is there friction on the object?
The answer is "yes", but this kind of friction is different from kinetic friction. Whereas kinetic friction acts to resist the motion of an object sliding across a surface, static friction is the force which keeps a motionless object from being pushed or pulled across a surface.

If a wooden block is at rest on a horizontal wooden surface, it is acted upon only by the normal force and the gravitational force.

If the surface is inclined by a small angle, θ, a component of the gravitational force acts downward along the surface of the board. The magnitude of this component is mg sinθ. If the block doesn't slide, then it is acted on by the static frictional force, fs, which exactly balances the mg sinθ component of the object's weight.


If the inclination of the surface is increased further, the static friction reaches a maximum strength. Unlike kinetic friction, which is roughly constant at low speeds, static friction varies to resist other forces on the block. The formula for static friction, therefore, gives only the maximum possible value of the static frictional force. The formula is as follows:

If the surface is inclined so much that the mg sinθ component of the object's weight exceeds fs max, then static friction is overcome, and the block begins to slide. At the angle where the block is just on the verge of slipping, fs max is equal to mg sinθ., and Newton's laws give us that

So the coeffiecient of static friction is equal to the tangent of the angle of inclination where the block is just on the verge of slipping.
Tanθ=.4
θ=21.8 degrees

2006-09-05 05:39:52 · answer #1 · answered by odu83 7 · 0 0

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