1. The problem statement, all variables and given/known data A man is dragging a 68 kg trunk up the loading ramp of a movers truck, the ramp has a slope of 20deg and the man pulls upward with a force F whose direction makes an angle of 30deg with the ramp. The coefficient of kinetic friction between trunk and the ramp is 0.28. What force F is necessary to accelerate the trunk up the loading ramp at a rate of 0.55m/s^2? Image: first one http://imageshack.us/photo/my-images/61/p6nj9.jpg/ 2. Relevant equations 3. The attempt at a solution My solution: I broke the gravity components into perpendicular and parallel so as a verticle equilibrium i got: mgcos20deg=Fn+Fsin30deg and as horizontal equilibrium i got: Fcos30deg=μFn+mgsin20deg I rearranged the horizontal equilibrium for F to find the Fn in the first equation like the following: F= ((μFn+mgsin20deg)/cos30deg) then: mgcos20deg=Fn+((μFn+mgsin20deg)/cos30deg)(sin30deg) By adding into all the given information i got 425.81N=Fn I plugged that into the friction formula Ff=μFn to find friction which i got was 119.2268N then i used this equation to use the acceleration given and find the F: Fnet=Fcos30deg-mgsin20deg-Ff ma=Fcos30deg-mgsin20deg-Ff F= 444.04 N But when i am checking the answer in the verticle equilibrium from before: mgcos20deg=Fn+Fsin30deg 626.21N=425.8N+222.02N 626.21N=647.82N what im i doing wrong? Pleas check and help me!