Question.
The workers and the platform together have a mass of 204kg and their combined centre of gravity is located directly above C. Given that the direction of the force exerted on the telescopic arm ABC at B by the single hydraulic cylinder BD is in the direction from D to B, find the force on the arm at A for the position when θ = 20°. You may neglect the weight of the telescopic arm in your calculation.
Answer
Let Fx and Fy be the forces at A towards +ve x- and y- directions.
Taking moments about B,
204*9.81*2.6cos20° + Fx*2.4sin20° + Fy*2.4cos20° = 0
=> 4889 + 0.821Fx + 2.255Fy = 0
=> Fx = - 5955 - 2.75Fy ... ( 1 )
Taking moments about D,
204*9.81*(5cos20° - 0.5) + 0.5Fy = 0.9Fx
=> 8402 + 0.5Fy = 0.9Fx=> Fx = 9336 + 0.556Fy ... ( 2 )
Taking moments about B,
204*9.81*2.6cos20° + Fx*2.4sin20° + Fy*2.4cos20° = 0
=> 4889 + 0.821Fx + 2.255Fy = 0
=> Fx = - 5955 - 2.75Fy ... ( 1 )
Taking moments about D,
204*9.81*(5cos20° - 0.5) + 0.5Fy = 0.9Fx
=> 8402 + 0.5Fy = 0.9Fx=> Fx = 9336 + 0.556Fy ... ( 2 )
Subtracting ( 1 ) from ( 2 ),
3.306Fy = - 15291
=> Fy = - 4625 N
Plugging Fy in ( 1 ),
Fx = - 5955 – 2.75 * (-4625) = 6764 N
F = √[4625)^2 + (6764)^2] = 8194 N
Angle with the horizontal
= arctan (4625)/(6764) = 34.4°.
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