A solid sphere of radius R and mass M is placed in a wedge as shown in the figur

Question

A solid sphere of radius R and mass M is placed in a wedge as shown in the figure. The inner surfaces of the wedge are frictionless.

pt. 1

1. FA = 2Mgsin(?)/sin(? + ?)

2.FA = Mgcos(?)/cos(? + ?)

3. FA = Mgcos(?)/sin(? + ?)

4. FA = Mgsin(?)/sin(? + ?)

5. FA = M g sin(?)/cos(? + ?)

6. FB = M g cos(?)/cos(? + ?)

Pt. 2

Determine the force exerted by the wedge on the sphere at the right contact point.

1.FB = M g sin(?)/ sin(? + ?)

2. FB = M g sin(?)/sin(? + ?)

3. FB = M g cos(?)/sin(? + ?)

4. FB = M g sin(?)/cos(? + ?)

5. FB = M g cos(?)/cos(? + ?)

6. FB = M g sin(?)/cos(? + ?)

Explanation / Answer

normal force due to surface A = N1

normal force due to surface B = N2

since the sphere is resting

so net vertical forces will be 0

so equating the vertical forces

N1 sin(alpha) + N2 sin(beta) = Mg

equating the horizontal force

N1 cos(alpha) = N2 cos(beta)

N2 = N1 cos(alpha) / cos(beta)

N1 sin(alpha) + N1 cos(alpha) sin(beta) / cos(beta) = Mg

N1 sin(alpha) cos(beta) + N1 cos(alpha) sin(beta) = Mg cos(beta)

N1 sin(apha + beta) = Mg cos(beta)

N1 = Mg cos(beta) / sin(alpha + beta)

force exerted by the wedge on the sphere at the left contact point = N1 = Mg cos(beta) / sin(alpha + beta)

now,

N1 = N2 cos(beta) / cos(alpha)

N2 sin(alpha) cos(beta) / cos(alpha) + N2 sin(beta) = Mg

N2 sin(alpha) cos(beta) + N2 sin(beta) cos(alpha) = Mg cos(alpha)

N2 sin(alpha + beta) = Mg cos(alpha)

N2 = Mg cos(alpha) / sin(alpha + beta)

force exerted by the wedge on the sphere at the right contact point = Mg cos(alpha) / sin(alpha + beta)

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