1. The problem statement, all variables and given/known data To move a large crate across a rough floor, you push on it with a force F at an angle of 21°, below the horizontal, as shown in Figure 6-21. Find the acceleration of the crate, given that the mass of the crate is m = 32 kg, the applied force is 344 N and the coefficient of kinetic friction between the crate and the floor is 0.54. http://www.webassign.net/walker/06-16alt.gif 2. Relevant equations a = Fnet/m Ffriction = Fnormal * muk 3. The attempt at a solution First I found the frictional force by (.54)(32 kg)(9.81 m/s2) = 169.5168 Now this is where I am stuck. Do you subtract that from the horizontal force? The horizontal force I got was 344Cos(21) = 321.15. Then I do 321.15 - 169.5168 = 151.63 and then divide that by the mass(32kg) and get 4.74 This is obviouley what you aren't supposed to do, so HELP!