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Old Oct 31st 2013, 03:11 PM   #1
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Friction

We have an inclined plane, on which is a cube. This cube does not move. Now, we decrease the angle of inclination of the plane.

Friction between the cube and the inclined plane:
a' increased
b' decreased
c' is the same
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Old Oct 31st 2013, 04:49 PM   #2
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Given that friction is proportional to the normal force between the block and inclined plane, you need to consider how the angle of incline affects the component of the block's weight that is normal (perpendicular) to the incline.
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Old Oct 31st 2013, 06:26 PM   #3
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Originally Posted by Peter View Post
We have an inclined plane, on which is a cube. This cube does not move. Now, we decrease the angle of inclination of the plane.

Friction between the cube and the inclined plane:
a' increased
b' decreased
c' is the same
Sorry, ChipB, I have to disagree. The cube is not moving so we are dealing with static friction...which is no larger than it needs to be to do the job (up to its maximum). In this case the force acting down the incline is the component of the weight force in that direction. The friction force is exactly that amount, but up the ramp (opposing the motion that the weight would cause.)

So what happens to the weight component down the incline as we decrease the angle of inclination?

-Dan
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Old Oct 31st 2013, 09:16 PM   #4
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Thanks Dan, you're right. It's a sly one!
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