The worker pushes downward and to the left on the box at an angle of 35°. The box has a mass of 135 kg. The box moves horizontally across the floor to the left at a constant velocity. The force applied by the worker is 2030 N∠215°. What is the value of the friction force? Fgravity · sin35° = Fgravityx Ffriction + Fgravityx = m · a f = μ · Fs sin35° = Fgravity · sin35° = Fgravityx m · g · sin35° = Fgravityx (135 kg)(9.81) sin35° = Fgravityx 759.6 N = Fgravityx Ffriction + (759.6 N) = (135 kg)(0 m/s2) Ffriction - 759.6 N = 0 N Ffriction = 759.6 N I know that this isn't correct, because my study guide is multiple choice. My options are: a. 1164 N b. 1324 N c. 1663 N d. 2489 N I'm really not sure what I'm missing or if I am going the wrong way with finding the value of the friction force. My guess is that c. 1663 N is correct, but that is only a guess. I'd like to be able to prove which answer it is. Thank you for any help!