(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A block with mass 10kg rests on a smooth, friction less ramp that is inclined at an angle of 45 degrees with the ground. How much force must be applied in the direction parallel to the ground to prevent the block from sliding down the ramp???

IT TURNS OUT THAT THE MAGNITUDE OF THE FORCE THAT MUST BE APPLIED IN THE DIRECTION PARALLEL TO THE GROUND IN ORDER TO PREVENT THE BLOCK FROM SLIDING DOWN THE RAMP IS EQUAL TO THE WEIGHT OF THE BLOCK (IN NEWTONS). Explain why this is the care. Will this always be the case for a block on a friction less ramp?

2. Relevant equations

I found the answer (please check) as 69.3N, the teacher gave us a follow up page saying the force should be equal to the weight of the block in newtons..Did I do this wrong?? The follow up page also asks to explain why the parallel force to the ground is equal to the weight of the block in Newtons. I dont get that.

3. The attempt at a solution

Weight = 10.9.8 = 98N

W vector <0,-98>

Vector u + Vector V = Vector W

F=-u

magnitude of F= magnitude of U

98 sin 45 = 69.3 N

A force of 69.3 N parallel to the plane will keep the weight from sliding.

3. The attempt at a solution

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# Homework Help: Mass on a friction less incline. What force is needed to stop the block from sliding.

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