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## Homework Statement

A 60 kg box is dragged along a rough but level floor by a rope at a 30

^{o}angle to the horizontal. The tension in the rope is 150 N, and the box is speeding up at 0.80 m/s

^{2}(along the floor).

(a) How large is the frictional force of the floor on the box?

## Homework Equations

F

_{frict}= u

_{k}*F

_{N}

F

_{N}= m*g - F*sin(theta)

## The Attempt at a Solution

I used F

_{N}= 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, F

_{N}= (60)(9.8) - (150)sin30 => F

_{N}= 588 - 75 =

**513**

From here, I know that F

_{frict}= u

_{k}*F

_{N}, but the only thing I can plug in is 513 for F

_{N}, 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?