A 90cm uniform lever has a load of 30N suspended at 15cm from one of its ends. If the fulcrum is at the center of gravity; the force that must be applied at its other end to keep it in horizontal equilibrium is?

By: |Published on: Nov 14|Categories: Physics JAMB| 0 comments

A. 15N
B. 20N
C. 30N
D. 60N

Correct Answer:

Option B – 20N

Explanation

Since the lever is a uniform one, its center of gravity will be acting half way (i.e 45cm)
∴from the principle of moments,
F x 45 = 30 x 30
∴ F = (30 x 30) / 45 = 20N

Users Also Read These:

SEE ALSO  The energy E of a photon and it's wavelength is related by Eλ = X. The numerical value of X is? [h = 6.63 x 10^-34Js, C = 3 x 10^8 ms−¹]

Copyright warnings! Do not copy.