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
Correct Answer: Option B
Explanation
Efficiency = ( frac{W(120N)INPUT}{WORK OUTPUT} times frac{100}{1} )
And for incline plane
V.R =( frac{1}{Sintheta} \
= 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