Question

The length of a bar of a screw jack is 35cm and it's pitch is 2cm, the jack is used to lift a vehicle of 880kg through a height of 15cm. If the efficiency of the screw jack is 40% calculate
I. Velocity ratio
Ii. Mechanical advantage
III. Effort applied
Iv. Work input
V. Work wasted
(g=10m/s², pi=22/7)

Answer 1

I. Velocity ratio = Distance moved by effort/Distance moved by load
Distance moved by effort = pitch of screw jack = 2cm
Distance moved by load = height lifted = 15cm
Velocity ratio = 2/15 = 0.133

II. Mechanical advantage = Load/Effort
Load = weight lifted = 880kg
Effort = force required to turn the screw jack
Efficiency of screw jack = 40%
Efficiency = (Load x distance moved by load)/(Effort x distance moved by effort) x 100%
40% = (880 x 0.15)/(Effort x 0.02) x 100%
Effort = 326.7N
Mechanical advantage = 880/326.7 = 2.69

III. Effort applied = 326.7N

IV. Work input = Effort x distance moved by effort
Distance moved by effort = pitch of screw jack = 2cm = 0.02m
Work input = 326.7 x 0.02 = 6.534J

V. Work wasted = Work input - Work output
Work output = Load x height lifted
Work output = 880 x 9.81 x 0.15 = 1297.32J
Work wasted = 6.534 - 1297.32 = -1290.8J (Negative value indicates that work is lost instead of being wasted due to inefficient process)