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?