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?
A
50N
B
100N
C
175N
D
210N
Verified Correct Answer: B.
Physical Principles & Numerical Solution
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
\(\text{but M.A } = \frac{load}{effort} = \frac{120}{\text{E}}\)
\(\text{therefore} \frac{120}{\text{E}}\) = 1.2
E = \(\frac{120}{1.2 }\) = 100N
Therefore, Effort up the plane = 100N
Source: Myschool Past Questions Archive (Record #41373)