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3 years ago

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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?

1 answer:

Mrac [35]3 years ago

8 0

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.

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