The tension in the rope An 800-N billboard worker stands on a 4.0 m scaffold supported by vertical ropes at each end. If the scaffold weighs 500...
Let AB be the scaffold . The length of AB = 4m.
Let G be the centre of gravity of the scaffold . Then assuming AB has uniformly dense along the scaffold, AG = AB =2m.
Let postion of the 800N worker be at X and AX =1meter.
To find the tension in the ropes at A and B.
We also presume the system forces is in equilibrium.
The forces in equilibrium are : The weight force of 800N (down ward) of the worker at X , the weight force of the scaffold 500N (down ward) Ta and Tb the tensions at A and B both upward. All forces are at equilibrium. Therefore,
Ta + Tb + 800N + 500N = 0 .....(1)
Taking moments about A,
Ta*0 + Tb*AB (clockwise) - 800* AX anti clock wise - 500* AG anticlockwise = 0 .
Tb*4m = 800*1 + 500*2 = 1800 Nm .
So Tb = 1800/4 N = 450 N upward
Ta =(800+500) - 450 = 850N upward