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?

50N

100N

175N

210N

Correct answer is B

Efficiency = $\frac{W(120N)INPUT}{WORK OUTPUT} \times \frac{100}{1}$

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

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