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]
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
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