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wolverine [178]
3 years ago
6

A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold welghs 198 N and is 3.2 m long.

what is the tension in each rope whien the 6oo-N worker stands 1.12 m from one end? smaller tension arger tension
Physics
1 answer:
iris [78.8K]3 years ago
5 0

Answer:

T_2 = 309 N

T_1 = 489 N

Explanation:

As we know that total tension in both the ropes is counter balancing the weight of scaffold and worker both

so here we will have

T_1 + T_2 = (m + M)g

now we have

T_1 + T_2 = 198 + 600 = 798 N

now we also know that net torque due to both tension force in the string with respect to the position of worker must be zero so that platform will remain in equilibrium and horizontal in position

so here we will have

T_1(1.12) + (198)(1.6 - 1.12) = T_2(3.2 - 1.12)

T_1+ 84.86 = 1.86 T_2

now from above two equations we will have

(1.86 T_2 - 84.86) + T_2 = 798

T_2 = 309 N

also we have

T_1 = 489 N

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The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
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<h3>Mass of the plane at the position of the rear wheels</h3>

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Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

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M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

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