A thin uniform rod of mass M is suspended horizontally by two vertical wires. On

Question

A thin uniform rod of mass M is suspended horizontally by two vertical wires. One wire is at the left end of the rod, and the other wire is 3/4 of the length of the rod from the left end. (a) Determine the tension (T) in each wire. (Use g for the acceleration due to gravity and M as necessary.) TL = TR = (b) An object is now hung by a string attached to the right end of the rod. When this happens, it is noticed that the rod remains horizontal but the tension in the wire on the left vanishes. Determine the mass m of the object. (Use M for the mass of the rod as necessary.) m =

Explanation / Answer

part A

left tension = Tl and right tension = Tr

Tl + Tr = Mg ...............1 ..............force balance

now do torque balance about center of maas of rod

Tl * ( L/2 ) = Tr * (L/4)

2* Tl = Tr .................2

from 1 and 2 we got

Tl = Mg / 3

Tr = 2Mg / 3

Part B

now again do torque and force balance about COM of rod

Tr + Tl = Mg + mg

but Tl = 0 so

Tr = mg + Mg

and torqeu balance

Tr * L/4 = mg * L/ 2

Tr = 2*mg

so

2*mg = mg + Mg

so

m= M ...................answer

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