Statics - sphere with string attached

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SUMMARY

This discussion focuses on solving two statics problems involving equilibrium conditions for rods and a sphere. The first problem requires proving that the coefficient of friction (mu) satisfies the inequality mu > 2/[tan(beta)+3tan(alpha)] for two uniform rods AB and BC. The second problem involves a uniform sphere of radius r and weight W, attached to a rough vertical wall via a string, and requires finding the angle of the string with the vertical and the tension in the string under the condition mu=h/2r. The solutions involve balancing forces and moments, emphasizing the importance of free-body diagrams.

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  • Understanding of equilibrium conditions in statics
  • Familiarity with free-body diagrams
  • Knowledge of frictional forces and coefficients
  • Basic principles of moments and torque
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  • Study the principles of static equilibrium in two-dimensional systems
  • Learn about free-body diagram techniques for complex structures
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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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