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The handle of a 22-kg lawnmower make a 35 degree angle with thehorizontal. If the coefficient of friction between lawnmower andground is .68, what magnitude of forceis required to push themower at constant velocity? Assume the force is applied in thedirection of the handle. Compare the force with the mowersweight.
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Answer #1

mass m = 22 kg

angle θ= 35 degrees

coefficient of kinetic friction μ = 0.68

frictional force f =  μ [ mg - F sinθ ]
where F = applied force
net force F ' = F cos θ - f
it moves with constant velocity.So, net force F ' =0
         F cos θ= f
       F cos 35 = 0.68 [ (22*9.8) - F sin 35 ]
       0.8191 F = 146.608 - 0.39F
                  F = 121.25 N
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Answer #2
For constant velocity, Fa(x-direction)=Ff

FF=μFn
FF=μ*(22kg*9.81m/s+Fa*sin(35))

Fa(x direction)=Fa*cos(35)

Fa=FF

Fa*cos(35)=0.68(22*9.81+Fa*sin(35))
Fa*0.8192=146.7576+Fa*0.3900
Fa*0.4292=146.7576
Fa=341.958N
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