Asked by ASIYA
a uniform pole,PQ 30m long of mass 4kg is carried by a boy at p and a man 8m away from Q.find the distance from p where a mass of 20kg should be attached so that the man's support is twice that of the boy,if the system is in equilibrium.[g=10m/s~-2]
Answers
Answered by
R_scott
total mass is 24 kg (20 + 4)
... so man should be supporting 16 kg , and boy supporting 8 kg
working from p
... center of mass of PQ is 15 m
... man is supporting at 22 m
... 22 * support = 15 * 4 kg ... support = 60/22 kg
... 20 kg * d = (16 kg - 60/22 kg) * 22 m
... so man should be supporting 16 kg , and boy supporting 8 kg
working from p
... center of mass of PQ is 15 m
... man is supporting at 22 m
... 22 * support = 15 * 4 kg ... support = 60/22 kg
... 20 kg * d = (16 kg - 60/22 kg) * 22 m
Answered by
ASIYA
i dnt understand pls explain it clearly
Answered by
Agana
Explain further for mine understanding,please
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