Question

a smooth circular cylinder of radi 1.5 m is lying in a triangular groove one side of which makes 15°and other 40°angle with the horizontal. find the reaction at the surface of contact if there is no friction and cylinder weighs 1000N​

Answer 1

Answer:

a smooth circular cylinder of radi 1.5 m is lying in a triangular groove one side of which makes 15°and other 40°angle with the horizontal. find the reaction at the surface of contact if there is no friction and cylinder weighs 1000N

Answer 2

Explanation:

Given A smooth circular cylinder of radius  1.5 meter is lying in a triangular groove as  shown which makes 15 degree angle and the other 40 degree angle with the horizontal. Find the reactions at the surface of  contact, if there is no friction and cylinder  weighs 1000 N

• The three forces R a, R b and weight must go through the centre of the cylinder.There will be no friction.

• So now by applying Lami’s theorem at point O we have,

• T / sin (180 – theta) = T / sin(180 – theta) = mg / sin 2 theta

• T / sin (180 – 40) = T / sin(180 – 15) = 1000 / sin 55

• T / sin 140 = T / sin 165 = 1000 / sin 55

•  R a / sin 140 = R b / sin 165 = 1000 / 0.819

• Ra / 0.642 = R b / 0.258 = 1000 / 0.819

• R a = 0.642 x 1000 / 0.819

•   R a = 783.88 N

• R b = 1000 x 0.258 / 0.819

• R b = 315.01 N

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