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
The correct answer is: B
Explanation
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