Question

8.A smooth circular cylinder of radius 1.5 meter is lying in a triangular groove, one side of which makes 15°

angle and the other 40° angle with the horizontal. Find the reactions at the surfaces of contact, if there is

no friction and the cylinder weights 100 N​

Answer 1

hello dear friend hope it help you xd 8

Answer 2

Answer:

The reactions at the surfaces of contact A and B are R1 = 12.25 N and R2 = 87.75 N, respectively.

Explanation:

From the above question,

They have given :

Let A and B be the two points of contact between the cylinder and the groove.

Let R1 and R2 be the reactions at A and B respectively.

Let θ1 = 15° and θ2 = 40°

We know,

R1 + R2 = 100 N ……(1)

Also, R1 Sin θ1 = R2 Sin θ2

R1 Sin 15° = R2 Sin 40°

R1 = (R2 Sin 40°)/Sin 15°

Substituting the value of R1 in equation (1),

R2 + (R2 Sin 40°)/Sin 15° = 100

R2 (1 + Sin 40°/Sin 15°) = 100

R2 = 100/(1 + Sin 40°/Sin 15°)

R2 = 87.75 N

R1 = 100 - 87.75

R1 = 12.25 N

Therefore, the reactions at the surfaces of contact A and B are R1 = 12.25 N and R2 = 87.75 N, respectively.

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