In the fig above,A lever of length 200m is used to lift a load of mass 180kg. The pivot at P is 20m from the load. what minimum force F must be applied at the end of the lever [g = 10ms-2]
Attached Figure / Diagram
A
9N
B
18N
C
20N
D
200N
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Applying the principle of moment
Clockwise moment = anti-clockwise moment.
But moment = force x perpendicular distance
F(200 - 20) = 1800 x 20
F = \(\frac{1800 \times 20}{180}\)
F = 10 x 20 = 200N
Source: Myschool Past Questions Archive (Record #41717)