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September 2, 2014

September 2, 2014

Posted by **Matthew** on Wednesday, October 7, 2009 at 8:44pm.

A 0.11- ball is placed in a shallow wedge with an opening angle of 130 degrees,for each contact point between the wedge and the ball, determine the force exerted on the ball. Assume the system is frictionless. Determine the Contact forces.

- Physics -
**bobpursley**, Wednesday, October 7, 2009 at 9:03pmI assume the point of the wedge is downward, so the forces acting on the ball is gravity.

If so, then the net force down is mg. Half that acts on each wall of the wedge.

Draw the figure. the normal force at the point of contact must be N, and using the vertical, the normal, and the side of the wedge, you should have tanTheta/2=N/W so

normal force=mg/2 * tanTheta/2

check my thinking.

- Physics -
**Matthew**, Wednesday, October 7, 2009 at 9:18pm\ 130o /

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C mg C

1 2

C= contact force #1 and #2.

Does this fit your logic?

- Physics -
**Aka**, Saturday, July 21, 2012 at 3:21amFirst of all it's in equilibrium; therefore, x and y force components are zero.

Secondly, it's almost a classic inclined plane problem, so the angle you should use is, I'll call it T, T = (90 - Angle/2).

Thirdly, the y component is Net Fy = NcosT - W/2 = 0. Remember, for this problem, use regular coordinates, not tilted axis. Because it is a mess if you use tilted axes on both halves. W (mg)= weight. it is W/2 because you take the half of the weight.

Last, solving for N yields: mg/2cosT = .11*9.81 /(2cos(90-65)) = .595

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