A 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N, directed 30 degrees below the horizontal. a.) Calculate the force of friction acting on the box.
b.) Calculate the normal force acting on the box.

Explanation:

Assuming the table is horizontal

As speed is constant, friction must exactly balance the horizontal component of the applied force

Ff = 50cos30 = 43 N

The  normal force must oppose both the weight of the box and the vertical component of the applied force.

N = 10(9.8) + 50sin30 = 123 N

i think the answer is a, hope this is right, that would be my best guess.

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