50N applied force is being pushed down on a 100kg block at an unknown angle (x). The coefficient of friction between the block and the ground is 0.4. regardless of the angle being used, what is the maximum elevation angle that will allow the 50N applied force to move the block? (image http://img296.imageshack.us/img296/7071/imagejq5.png)
2006-11-04 10:51:25 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
http://img296.imageshack.us/img296/7071/imagejq5.png
thats the picture
2006-11-04 11:53:34 · update #1
0.4*(100kg*g+F*sin(x)) = F*cos(x). ignoring any rotational component
2006-11-04 11:14:48 · answer #1 · answered by arbiter007 6 · 0⤊ 0⤋
Well, I wasn't able to see the image because the link isn't complete, but if i'm not wrong you have to push upward in order to move the block, because the friction created by the weight of the block is bigger than the 50 N force, I would dare saying that not even using the 50 N force completely upwards you would be able to move that block.
2006-11-04 11:45:46 · answer #2 · answered by mensajeroscuro 4 · 0⤊ 0⤋
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2016-12-28 12:56:02 · answer #3 · answered by ? 3 · 0⤊ 0⤋