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A uniform bar of mass m and length L is ...

A uniform bar of mass m and length L is horizontally suspended from the ceiling by two vertical cables A is connected at a distance of `L/4` from the left end of the bar. Cable B is attached at the far right end of the bar. What is the tension in cable A?

A

1/4Mg

B

1/3Mg

C

2/3Mg

D

3/4Mg

Text Solution

Verified by Experts

The correct Answer is:
C
`sumtau_B=T_A(3/(4L))-Mg(1/(2L))=0` Therefore`T_A=((MgL)/2)/((3L)/2)=((MgL)/2)/((4)/(2L))`
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