Question

A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer of mass 79 kg stands at a point 1.9 m from one end. (a) What is the tension in the cable closer to the painter? (b) What is the tension in the cable further from the painter?

Answer 1

Explanation:

The given data is as follows.

     $m_{1}$ = 57 kg,        $m_{2}$ = 79 kg

      $l_{1}$ = 6.5 m,         $l_{2}$ = (6.5 - 1.9) m = 4.6 m

(a)  The sum of torque ends about far end is as follows.

    $m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1}$ = 0

     $57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5$ = 0

                     T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b)  Now, we will calculate the sum about close ends as follows.

$m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}$    

  $57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5$ = 0

                            T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.