A screw jack with a pitch of 0.2 cm and a handle of length 50 cm is used to lift a car of weight 1.2×104N.
If the efficiency of the screw is 30%, find:
(i) The velocity ratio and mechanical advantage of the machine
(ii) The effort required to raise the car
Solution:
Data Given:
Pitch, p=0.2cm
Radius, R=50cm
Load, L=1.2×104N
Efficiency, η=30%
Step 1: Calculate Velocity Ratio (V.R)
V.R=p2πR=0.22×3.14×50
V.R=0.2314=1570
Step 2: Use Efficiency Formula to find M.A
η=V.RM.A×100%
10030=1570M.A
M.A=0.3×1570=471
Step 3: Use M.A to find Effort
M.A=EffortLoad
471=Effort1.2×104
Effort=4711.2×104
Effort≈25.48N
Final Answer:
Velocity Ratio = 1570
Mechanical Advantage = 471
Effort ≈ 25.48 N