A particle of weight 120N is placed on a plane inclined at 300 to the horizontal. If the plane has an efficiency of 60% to the floor what is the force requi...

A particle of weight 120N is placed on a plane inclined at 300 to the horizontal. If the plane has an efficiency of 60% to the floor what is the force required to push the weight uniformly up the plane?

Answer Details

Efficiency = $\frac{W\left(120N\right)INPUT}{WORKOUTPUT}\times \frac{100}{1}$

And for incline plane

V.R =$\frac{1}{Sin\theta }=V.R=\frac{1}{Sin30}=\frac{1}{0.5}=2\text{therefore}\frac{60}{120}=\frac{M.A}{2}=M.A=\frac{120}{100}=1.2\text{but M.A}=\frac{load}{effort}=\frac{120}{E}\text{therefore}\frac{120}{E}=1.2=E=\frac{120}{1.2}=100N$

Therefore Effort up the plane = 100N