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The diagram shows a light rod AB of length 4a rigidly joined at B to a light rod BC of length 2a so that the rods are perpendicular to each other and in the same vertical plane. The Centre O of AB is fixed and the rods can rotate freely about O in a vertical plane. A particle of mass 4m is attached at A and a particle of mass m is attached at C. The system rests in equilibrium with AB inclined at an acute angle θ to the vertical as shown. By taking moments about O, find the value of θ.

Moment = Force × distance [perpendicular to the force measured from the pivot]We know that because it is in equilibrium, then the sum of clockwise and anticlockwise moments = 0.mass1 × gravitational field strength × perpendicular distance AO − mass2 × gravitational field strength × (perpendicular distance OB + perpendicular distance BC) = 04mg × 2a × sinθ − mg × (2a × sinθ + 2a × cosθ) = 0mg[ 4 × 2a × sinθ − 2a × sinθ − 2a × cosθ ] = 02amg(4sinθ − sinθ − cosθ) = 02amg(3sinθ − cosθ) = 03sinθ − cosθ = 03sinθ = cosθ3sinθ/cosθ = 13tanθ = 1tanθ = 1/3θ = tan^-1(1/3) = 0.322 rad
Sam P.
05 May 2016
08 June 2016
drderekk helpfully has a diagram for the same setup but with the long rod length of 5a instead of 4a and a mass of 3m instead of 4m at point A.However drderekk has taken moments about point B which does not work because this is answering a different question where the pivot is at B instead of O.Please see my answer below for the correct solution.
Sam P.
10 June 2016
Leonora your solution is incorrect. You have correctly stated that a moment = force × perpendicular distance [from the pivot] but you have not calculated the perpendicular distance to the force mg correctly (there is a term missing).Again, please see my answer below for the correct solution.
Sam P.
01 September 2016
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