M*g = 27 * 9.8 = 264.6 N. = Wt. of the lawnmower.
Fn = 264.6 + Fap*sin36=264.6 + 0.588Fap
Fk = u*Fn = 0.66*(264.6+0.588Fap) =
174.64 + 0.388Fap
(Fx-Fk) = M*a
(Fap*Cos36-(174.64+0.388Fap) = M*a
(0.81Fap-174.64-0.388Fap = M*0
0.42Fap - 174.64 = 0
0.42Fap = 174.64
Fap = 416 N. = Force applied.
The handle of a 27kg lawnmower makes a 36∘ angle with the horizontal.
If the coefficient of friction between lawnmower and ground is 0.66, what magnitude of force, applied in the direction of the handle, is required to push the mower at constant velocity?
Express your answer using two significant figures.
Compare with the mower’s weight.
Express your answer using two significant figures.
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