I 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.
Please help me out, I can't figure out the process for answering this question and my TA is of no help,
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.
3 answers
\ 130o /
\ /
\ /
\ /
\O/
/|\
C mg C
1 2
C= contact force #1 and #2.
Does this fit your logic?
\ /
\ /
\ /
\O/
/|\
C mg C
1 2
C= contact force #1 and #2.
Does this fit your logic?
First 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
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