Equilibrium of Two People Holding Uniform Beam

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The discussion focuses on calculating the forces exerted by two individuals holding a uniform beam of length 7.6 meters and weight 450 Newtons. The first person applies force F1 at a distance of 1 meter from the left end, while the second person applies force F2 at a distance of 2 meters from the right end. The equilibrium equations established are sum of forces (F1 + F2 = 450) and sum of torques (4.6F2 - 1260 = 0), leading to definitive calculations of the forces exerted by each individual.

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two people are holding a unifrom beam of length 7.6m and weight 450N. the first person is holding the beam 1 m from the left end the second is holding the beam 2m from the right end determine the force the each person exerts on the beam.

choosing the axis as the left most person who exerts F1

sum F = F1 + F2 -450=0
sum torque = 4.6F2-2.8*450 =0
 
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hi nameVoid! :wink:

that looks ok :smile: …

what's worrying you about it? :confused:​
 
thanks i was unsure where to take the axis
 
it doesn't matter …

take the axis anywhere (even somewhere really stupid! :biggrin:), and it'll still work :smile:

(though it make the maths easier if you always take the axis through one of the forces, in this case one of the two people or the centre of mass :wink:)
 

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