A 19.0 kg box sits on a frictionless ramp with a 14.9° slope. The movers pull up a rope attached to the box to move it up. If the rope makes an incline with the ramp that is 42.0° to the horizontal, what is the smallest force F the mover must use to move the box up the incline?
Newton's 2nd Law
The Attempt at a Solution
I was already given the answer to this as 53.8 N, but I can't get there (though I get very close)
I broke it down into 3 forces; the gravitational force, the normal force, and the force the mover exerts on the box. Getting the value 186.2 N for the value of Fgy, I obtained 47.88 N for Fgx by utilizing Fgy divided by cos∅ and then multiplying that value by sin∅. The F exerted by the mover along the x should equal that value of 47.88. Using that value, 47.88 / cos 56.9° (The angle the rope makes with the rotated axis I used) I get a monstrous value of 87.68 N. Any tips or pointers would be much appreciated.