The tension in the rope An 800-N billboard worker stands on a 4.0 m scaffold supported by vertical...

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