A 40-N crate rests on a rough horizontal floor. A 12-N horizonatl force is then applied to it. If the coefficients of friction are s = 0.5 and k = 0.4, the magnitude of the frictional force on the crate is: what i did was : 40*0.5=20 , 40*0.4=16 20+16=36 40/12=3.3 36/3=12 how to solve this problem? could someone help? thank you.