A cylinder of weight 500N is kept on two inclined planes as shown in the figure. Determine the reactions at the contact points A & B.

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written 3.9 years ago by

teamques10 ★ 69k

modified 3.4 years ago by

yashbeer ★ 11k

engineering mechanics

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written 3.9 years ago by

teamques10 ★ 69k

Let C be the centre and A, B be the contact points.

$Reaction\ forces\ R_A, R_B\ and\ weight\ 'W' are\ under\ equilibrium\, at\ centre\ point\ C.$

$Hence, \dfrac {R_A} {sin(180-30)} =\dfrac {R_B} {sin(180-50)} = \dfrac {500} {sin(30+50)}$

$Hence, \dfrac {R_A} {sin(150)} =\dfrac {R_B} {sin(130)} = \dfrac {500} {sin(80)}$ …

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