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 2/3 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.)

(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

system is in equilibrum so net torque will be 0

torque about left end = Mg * L / 2 - TR * 2L / 3

Mg * L / 2 - TR * 2L / 3 = 0

Mg * L / 2 = TR * 2L / 3

3Mg / 4 = TR

TR = 3Mg / 4

torque about point where right wire is attached = TL * 2L / 3 - Mg * (2L / 3 - L / 2)

TL * 2L / 3 - Mg * (2L / 3 - L / 2) = 0

TL * 2/ 3 - Mg / 6 = 0

TL = Mg / 4

after hanging new mass

torque about point where right wire is attached = Mg * (2L / 3 - L / 2) - mg * L / 3

Mg * (2L / 3 - L / 2) - mg * L / 3 = 0

Mg * (2L / 3 - L / 2) = mg * L / 3

M / 2 = m

m = M / 2

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