Uniform rod resting against rough vertical wall supported by a string.

[kozor]

Homework Statement

Uniform rod AB with end B resting against rough vertical wall. Coefficient of friction between wall and rod is μ.

Rod is 2m long and has mass 3kg.

Rod is kept in limiting equilibrium by a light inextensible string, one end of which is attached to the end A of the rod and the other to point C on the wall 2m above B. Angle ABC=120°.

The end B of the rod is about to slip down the wall.

1. By taking moments about C, show that the normal reaction at B is approx. 12.7N.

2. Find the value of μ.PS: A labelled diagram would be very helpful! Thanks!

Homework Equations

n/a

The Attempt at a Solution

Have been able to sketch what I think are all the forces acting on the rod, but does the normal reaction + friction + T-resolved = downward force of 2g-resolved? Do I take the weight of the rod to act from the centre?

 

[paisiello2]

PS: A labelled diagram would be very helpful! Thanks!

Yes, it would! Can you provide one?

 

[kozor]

No worries.

http://imageshack.com/a/img836/6874/gykt.png

 

[dauto]

Yes, unless otherwise stated, its safe assume the rod is uniform in which case its center of gravity is at the center

 

[HalfLostAndSearching]

I would approach this problem by first identifying all the forces acting on the rod and using the principles of equilibrium to solve for the unknown values.

For the first part, taking moments about point C, we can set the clockwise moments equal to the counterclockwise moments:

Clockwise moments: 0.5m x 2g x sin(120) = 2mg
Counterclockwise moments: μ x n x 2m = μn x 2m

Setting these two equal, we can solve for n, the normal reaction at B, which is approximately 12.7N.

For the second part, we can use the principle of equilibrium to solve for the coefficient of friction μ. Since the rod is in limiting equilibrium, the sum of the forces in the horizontal direction must be equal to 0. This means that the friction force must be equal and opposite to the horizontal component of the tension in the string.

Using trigonometry, we can calculate the horizontal component of the tension to be T x cos(120). Setting this equal to the friction force, we can solve for μ using the equation μ = Ff/n.

I would also recommend drawing a free body diagram to help visualize the forces acting on the rod and to make it easier to solve the problem. Overall, the key is to use the principles of equilibrium and to carefully consider all the forces acting on the rod.