# Find the value of tension just sufficient to achieve limiting equilibrium in the following case: A heavy ring of mass 5kg is threaded on a fixed rough horizontal rod. The coefficient of friction...

Find the value of tension just sufficient to achieve limiting equilibrium in the following case:

A heavy ring of mass 5kg is threaded on a fixed rough horizontal rod. The coefficient of friction between the rod and the ring is 0.5. A string is attached to the ring and pulled downwards with a force acting at a constant angle of 30 degrees to the horizontal. The magnitude of the force is T newtons , and is gradually increased from zero.

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Here we have a ring threaded on a fixed rough horizontal rod with the coefficient of friction between the rod and the ring being 0.5

A string is tied to the ring and pulled down with a tension equal to T at 30 degrees to the horizontal. The force on the ring due to the tension can be divided into two components, one in the vertical direction acting downwards and another in the horizontal direction.

The total force on the ring acting downwards is 5*g + T* sin 30.

The force of friction is given by the product of the coefficient of friction and the normal force = (5*g + T* sin 30)*0.5.

The force acting on the ring trying to move it in the vertical direction is equal to T*cos 30. This is opposed by the force of friction in the opposite direction equal to (5*g + T*sin 30)*0.5

When the two forces are equal the equilibrium is limiting,

T*cos 30 = (5*g + T*sin 30)*0.5

=> T*cos 30 - 0.5*T*sin 30 = 5*g

=> T (cos 30 - 0.5*sin 30) = 5*9.8

=> T = 5*9.8/ (cos 30 - 0.5*sin 30)

=> T = 49/ (sqrt 3 / 2 - 0.25)

=> T = 79.542

Therefore the tension that has to be applied to reach limiting equilibrium is 79.542 N.

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