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NeX [460]
1 year ago
15

calculate the tension in the horizontal rope. (the horizontal and the vertical ropes are not connected to each other. they are b

oth independently attached to the end of the boom.)
Physics
1 answer:
Brilliant_brown [7]1 year ago
8 0

There is a tension of 7,019.4 N on the horizontal rope.

The pulling force that travels axially along a rod's ends, a string, a cable, a chain, or another similar device is known as tension.

The action-reaction pair of forces that are acting at the ends of the element are also known as tension.

The tension is measured in Newtons. On the horizontal rope, there is tension.

The following formula is used to determine the horizontal rope's tension:

Utilize the torque concept.

Mg (L/2) sin + mg M + 2mg sin/cos = T (M + 2m) L sin = T(L/2) cos 660 g tan = T let (This value should be included in the question)

(M + 2m)

GTAN=TTAN= (74.9 + 2 x 122) (9.8) (tan 66)

T = 7,019.4 N The tension in is 7,019.4 N.

The complete question is - An object with a mass of m = 122 kg is suspended by a rope from the end of a uniform boom with a mass of M = 74.9 kg and a length of l = 8.77 m. The end of the boom is supported by another rope which is horizontal and attached to the wall as shown in the figure.

Calculate the tension in the horizontal rope. (The horizontal and the vertical ropes are not connected to each other. They are both independently attached to the end of the boom.)

Learn more about Tension here-

brainly.com/question/6359509

#SPJ4

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Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

              95 kg skydiver: 12.3 m/s and 386.2 s

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mg=\frac{1}{2}C \rho Av_{T}^{2}

v_{T}=\sqrt{\frac{2mg}{\rho CA} }

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ρ is density of the fluid in kg/m³

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A is area of the object in the fluid in m²

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The 52kg skydiver has terminal velocity of:

v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }

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The 95kg skydiver's terminal velocity is

v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }

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The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

The time each one will reach the floor will be:

52 kg at 9.09 m/s:

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95 kg at 12.3 m/s:

t=\frac{4750}{12.3}

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

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