Question

Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 805 N and stands 1.00 m from the left end. Two meters from the left end is the 500 N washing equipment. Joe is 0.500 m from the right end and weighs 820 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable

Answer 1

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

Weight of Joe = 820 N and stand 0.500 m from the right end

so the force on the left cable will be;

$T_{left$ = $\frac{1}{3m}$[ (805 N)( (3-1) m) + ( 395 N )( $\frac{3}{2} m$) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]

$T_{left$ =  $\frac{1}{3m}$[ 1610 + 592.5 + 500 + 410 ]

$T_{left$ =  $\frac{1}{3m}$[ 3112.5 ]

$T_{left$ =  1037.5 N

$T_{left$ =  1.038 kN

Therefore, the forces on the left hand side is 1.038 kN

On the right hand side;

$T_{Right$ =  ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N

$T_{Right$ =  2520 N - 1037.5 N

$T_{Right$ =  1482.5 N

$T_{Right$ =  1.483 kN

Therefore, the forces on the right hand side is 1.483 kN