A body lifts a load of 400N through a vertical height of 5m in 10 s by a fixed pulley by applying an effort of 480N at the other end.
i) Find the velocity ratio.
ii) Find the mechanical advantage
iii) Find the efficiency.
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Answers
L = 400 kgf,
dL = 2 m,
t = 5min 10sec = 310sec
E = 48 kgf.
(i) Mechanical advantage (M.A.) = L/E = 400/48 = 8.33
(ii) If the effort moves a distance d downwards, the load also moves a distance d upwards. So velocity ratio (V.R.) = d/d = 1,
Efficiency = M.A./V.R. = 8.33/1 = 8.33(or 83.3%).
The efficiency of the pulley is not 100% because some energy is wasted in overcoming the friction in the pulley bearings.
(iii) The energy gained by the load in 5s = Load x Displacement of load in 5s
= 400 kgf x 2 m = 800 kgf x m.
(iv) Power developed by the boy = Effort x Displacement of effort/Time = 48 kgf x 2m/5s
= 19.2 kgf x ms-1
The answers to the above question are given below -
Given: Force = 400N
Vertical height = 5m
Time = 10 seconds
Force applied to the pulley = 480N
To find: (a) Velocity ratio
(b) The mechanical advantage
(c) Efficiency
Solution:
(A) Velocity ratio is the ratio between the displacement of effort and displacement of load.
V.R = displacement of the effort/displacement of load.
V.R = De/Dl
Since the distance traveled by load and effort is the same, therefore the velocity ratio is 1.
(B) Mechanical advantage is the ratio between load and effort is called mechanical advantage (M.A).
M.A = Load/Effort
= 400/480
= 5/6
Mechanical advantage is 5/6.
Since it is a ratio therefore it has no unit.
(C) Efficiency of the machine is defined as the ratio of output and input.
Efficiency = 480/400 = 6/5
Therefore, the efficiency is equal to 6/5.