Frictional Force

[soby10]

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From TPR workbook.
"A person is pulling a block of mass m with a force equal to its weight directed 30 degrees 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 degrees above the horizontal plane across the same rough surface, what is the friction on the block? (mu k is the coefficient of kinetic friction across the surface)".

Answer is 3f.
I don't know how they got that answer. I know that the x component of the weight force is mg*sin*30 degrees pushing block down and Fapplied is the same value in opposite direction. Fk is the opposing force against Fapplied and it is mgsin30. Is Fn decreased by mgsin30? Overall what is the equation needed to solve for friction force?

 

[enervate]

You'd need to calculate sin(30) = 0.5 and simplify the equations for f in both situations. You should get f_a = (0.5) * mg * mu_k and f_b = (1.5) * mg * mu_k.

edit: Recall that the equation for the force by friction is essentially F_y * mu_k.

In the first situation, F_y = [mg - mg * sin(theta)]. In the second situation, F_y = [mg + mg * sin(theta)].