A rigid rod with mass 20 kg and length L is attached to a wall by a frictionless

Question

A rigid rod with mass 20 kg and length L is attached to a wall by a frictionless pivot as shown. A mass of 5 kg hangs from the rod at (3/4)L, and a string is attached at L. The center of mass of the rod is at L/2. The angle theta is 26o, and L is 2 m.

Determine the magnitude of the tension T.

Find the magnitude and direction of the reaction force acting at the pivot.

Suppose the string with tension T suddenly broke. Find the torque about the pivot at the instant it snapped.

Any help would be really appreciated!

Explanation / Answer

vertical forces :

Tsin + Ncos = mg + Mg

horizontal force provides the normal reaction.

=> Tcos = Nsin

say N, reaction force acting at an angle to the vertical

now the net torque = 0

=> Tsin(L) - mg(3/4)L -Mg(L/2) = 0

=> T = 313.7N

and solving for

we get = 68.25 from vertical

and magnitude of N = 303.56 N

at the instant snapped : net torque : 20(1) + 5(0.75*2) = 275N-m

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