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April 17, 2014

April 17, 2014

Posted by **L** on Friday, June 11, 2010 at 10:45am.

8 hours ago - 4 days left to answer.

Additional Details

It looks as though the 30 cm length coverst the length of the wire from the wall to the surface of the sphere. When I worked it out I added the length of the radius to it.

- physics -
**drwls**, Friday, June 11, 2010 at 11:07amWhere is the wire attached to the ball? If the wall contact point is frictionless, the wire attached to the ball must, if extended, pass thtough the center of the ball; otherwise the string will apply an unbalanced torque about the center.

The angle the string makes with the wall is given by

sin A = R/(R + 30) = 32/62

A = 31.1 degrees

Now apply a moment balance about the string contact point at the wall. The string tension applies no moment there since it passes through the point. If F is the contact force at the wall,

M g *32 = F * (30 + 32)cos 31.1

F = 45*9.8/[0.62 cos 31.1)= 830 N

You can get the string tension T from a hotizontal force balance.

F = T sin 31.1

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