Statics - sphere with string attached

AI Thread Summary
The discussion centers on solving two statics problems involving equilibrium conditions. The first problem involves two uniform rods in equilibrium, requiring the proof that the coefficient of friction (mu) is greater than 2 divided by the sum of the tangents of angles beta and alpha. The second problem involves a uniform sphere attached to a wall by a string, where the angle of the string with the vertical and the tension in the string need to be determined based on the coefficient of friction. Participants emphasize the importance of balancing horizontal and vertical forces and taking moments about strategic points to solve these problems. Proper diagramming is also highlighted as crucial for visualizing the forces involved.
tyche
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I'm really stuck on these two problems:

1) Two equal uniform rods AB BC are smoothly jointed in B and they're in equilibrium. The end of C lies on a rough horizontal plane and the end of A is freely pivoted at a point above the plane. If alpha and beta are the angles of CB and BA with the horizontal, prove that

mu > 2/[tan(beta)+3tan(alpha)]

2)A uniform sphere of radius r and weight W has a ligjht inextensible string attached to a point on its surface, while the other end is attached to a rough vertical wall. The sphere is in equilibrium touching the wall at a distance h below the point of attachment to the wall and is about to slip.
a)If coefficient of friction=mu find the angle of the string with the vertical
b)If mu=h/2r show that the tension in the string is W[Sqrt(1+mu^2)]/2mu


I've tried every combination of forces that came in my mind, with no result.
Of course I've balanced, taken the moments, etc..
Can someone help me? Thanks!
 
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In all these kinds of statics problems, the way to proceed is to equate the sum of the horizontal and vertical forces individually to zero; and to equate the moment of all the forces about a conveniently chosen point to zero.

Problem 2:
Let F be the frictional force acting upward at the point of contact, and N the normal reaction acting toward the centre at this point. T is the tension in the string, and makes an angle ‘b’ with the vertical.

Horizontal forces: N=Tsin b .
Vertical forces: W=F+Tcos b.

Taking moment about the point where the string is attached to the wall, Nh=Wr.
Also, if k is the co-eff of friction, then, F=kN.

To find tan b, divide 1st eqn by 2nd, and plug in values of F and N in terms of W, r and h.
To find T, square and add the 1st and 2nd to eliminate b and get T^2.

Problem 1:
For this problem, same approach. The weights of the rods act at their mid-pts. There are two normal reactions at the points A and C, and a horizontal force of friction F at point C. It looks scary but is not. Draw a proper diagram.
 
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