The NET force up the incline must be
Fnet = M*a = 5.05*4.3 = 21.72 N
This equals the sum of of the applied force component up the ramp (Fap*cos36.5), minus the friction force (M*g*muk*cos36.5-Fap*sin36.5muk)) and minus the weight component down the ramp (M*g*sin36.5).
Solve for Fap
A 5.05kg sits on a ramp that is inclined at 36.5 degrees above horizontal.The coefficient of kinetic friction between box and ramp is mu k= 0.25. What horizontal force is required to move the box up the incline with a constsant acceleration of 4.3m/s^2?
1 answer
1 answer