Incline Plane & Friction Question

  • Thread starter bchung606
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Question says: A person is pulling on a block of mass m with a force equal to its weight directed 30° above the horizontal plane across a rough surface, generating a friction f on the block. If the person is now pushing downward on the block with the same force 30° above the horizontal plane across the same rough surface, what is the friction on the block?

Solution says: The friction f on the block is represented by the formula f =μkN, where N is the normal force acting on the block. When the force is applied 30° above the horizontal, N = mg - mg sin30° Since sin30° is 0.5, N = mg - 0.5mg = 0.5mg. Substituting N into the formula for friction, it becomes f1 = 0.5μk. When the force is applied 30° below the horizontal, N = mg + mg sin30° = mg + 0.5mg = 1.5 mg. Substituting N into the formula for friction, it becomes 1.5μk = 3f1.


I think I understand what the book is saying logically, but I don't understand how N = mg - mgsinθ or that N = mg + mgsinθ. N should equal mgcosθ. Can someone help me understand this part?
 

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Doc Al
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I think I understand what the book is saying logically, but I don't understand how N = mg - mgsinθ or that N = mg + mgsinθ. N should equal mgcosθ. Can someone help me understand this part?
First, realize that this is a horizontal plane, not an inclined plane. To find the normal force, just add up the vertical forces acting on the block. They must add to zero.

In the case where the force is applied at an upward angle, what forces act on the block and what are their vertical components?
 
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