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

Step-by-step 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 (l_(1))/(2) + m_(2)g * l_(2) - T * l_(1)` = 0

`57 * 9.81 * (6.5)/(2) + 79 * 9.81 * (6.5 - 1.9) - T * 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 (l_(1))/(2) + m_(2)g * (1.9) - T * l_(1)`

`57 * 9.81 * (6.5)/(2) + 79 * 9.81 * (1.9) - T * 6.5` = 0

T= 506 N

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