Two smooth uniform spheres of radius 4cm and mass 5kg are suspended from the same point A by light inextensible strings of length 8cm attached to their surfaces. There spheres hang in equilibrium, touching each other. What is the reaction between them?

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From how i understand this, the reaction force should be indeterminable.

Resolving horizontally: Tsinø + R (left hand side) = Tsinø + R (right hand side)

So why is the reaction force then specifically the sine of the tension in one of the strings?

If the two spheres were both on the ground and touching each other, there would still be a reaction between them, so i can't understand why the tension in the string is a cause of the reaction force?

  • $\begingroup$ Since T is not dimensionless, the reaction cannot have the form $sin(T)$ $\endgroup$ – Dr Chuck Apr 22 '17 at 18:21

The net force on each sphere is zero because each is in equilibrium. You seem to be satisfied with the balance of vertical forces between tension in the string and the weight of the ball : $T\cos\theta=mg.$

Horizontally the forces on each ball are horizontal component of tension and normal reaction. These are equal and opposite : $T\sin\theta=R$.

You seem to be trying to balance horizontal forces on both balls at the same time. Then you have external forces $T\sin\theta$ to the right and $T\sin\theta$ to the left. Unless the tensions $T$ are different or $\theta$ is different, that does not tell us anything new.

The normal reactions $R$ are internal forces. Together they always cancel out because they are always equal and opposite pairs. To find normal reaction we have to consider the forces on each ball individually. The normal reaction from ball A is then an external force acting on ball B.

  • $\begingroup$ OK, i'm pretty sure I get your explanation. Additionally, how would i know whether or not the reaction force on each is horizontal? What if it were acting at an angle? $\endgroup$ – Logan545 Apr 22 '17 at 18:46
  • $\begingroup$ @Logan545 reaction force is always​ "normal" ie at right angles to the surface where the two objects touch. In the case of two spheres this is along the line joining their centres, which (as they are on string the same length) is horizontal. $\endgroup$ – IanF1 Apr 22 '17 at 18:50
  • $\begingroup$ @Logan545 The spheres are described as "smooth." This means there is no friction parallel to the surface. By definition the only contact forces are normal to the surface. If the spheres were "rough" this would be a much more difficult situation to analyze. $\endgroup$ – sammy gerbil Apr 22 '17 at 18:59
  • $\begingroup$ Ok i see. So as an extension, if they were rough, would getting the reaction force be a simple matter of resolving iy horizontally and vertically and maybe taking moments? Or would there me more than that to it? $\endgroup$ – Logan545 Apr 22 '17 at 19:20
  • $\begingroup$ Yes, moments would be required. With only 2 balls I think this is solvable using the rigid body model for the spheres. But some cases (more balls and strings) are indeterminate - eg Reaction forces in pyramid stacking of steel coils. Then you have to allow each body to deform. $\endgroup$ – sammy gerbil Apr 22 '17 at 20:07

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