A 60 kg box is dragged along a rough but level floor by a rope at a 30o angle to the horizontal. The tension in the rope is 150 N, and the box is speeding up at 0.80 m/s2 (along the floor).
(a) How large is the frictional force of the floor on the box?
Ffrict = uk*FN
FN = m*g - F*sin(theta)
The Attempt at a Solution
I used FN = m*g - F*sin(theta) to calculate the normal force, which is needed because it is a component of the equation to find frictional force. So, FN = (60)(9.8) - (150)sin30 => FN = 588 - 75 = 513
From here, I know that Ffrict = uk*FN, but the only thing I can plug in is 513 for FN, which still leaves me with two unknowns.
Can anyone be so kind as to tell me the next step? Am I missing something? Looking at the problem completely wrong?