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Aliun [14]
3 years ago
5

A window washer is standing on a scaffold supported by vertical ropes at each end. The scaffold weighs 200. N and is 3.00 m long

. What is the tension in each rope when the 700. N worker stands 1.00 m from one end?

Physics
1 answer:
seropon [69]3 years ago
5 0
<h2>Tension in ropes are 333.33 N and 566.67 N.</h2>

Explanation:

Refer the given figure showing the arrangement.

For having stability we have torque at A and B are zero.

        Sum of Torques at A = 0

        700 x 1 + 200 x 1.5 - Q x 3 = 0

                               3Q = 1000

                                  Q = 333.33 N

        Sum of Torques at B = 0

        200 x 1.5 + 700 x 2 - P x 3 = 0

                               3P = 1700

                                  P = 566.67 N  

Tension in ropes are 333.33 N and 566.67 N.

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Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

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Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

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v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

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You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

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You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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