Question

A crate on a floor has an applied force 15 degrees below horizontal of 952N The crate moves with constant velocity. The coefficient of kinetic friction is 0.15. What is the mass of the crate ?
bobpursley
break the force into vertical and horizontal components. The horizonal component moves the crate, and the upward component lessens friction.

fvrtical=952*sin15
friction force=.15(mg-952sin15)

but at constant velocity, friction force must equal the horizontal component.

.15(mg-952sin15)=952cos15
solve for mass m.

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