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Rus_ich [418]
2 years ago
4

A (A+60.0) kg construction worker is standing on an 18.0 kg plank. The plank is (3.50+B) m long and it is suspended by vertical

cables at each end. If the worker stands 2.80 from the right end of the plank, what is the tension in right cable? Give your answer in newtons (N) and with 3 significant figures.
Physics
1 answer:
adoni [48]2 years ago
3 0

Answer:

The force at right side is 1058N

Explanation:

The concept that we need to use to give a solution to this problem are the Static Equation where the force and torque do not experience an acceleration.

Our values are:

m_1 = 60Kg\\m_2 = 18Kg\\L = 3.5m\\d = 2.8m

We have two tension from the wire, at left and right, then making sum we have

\sum F = 0

T_L+T_R-mg=0

T_L+T_R = (60)(9.8)

T_L+T_R = 588N

We can do now a sum of moments at right side, then

\sum T = 0

mg*2.8+T_L*3.5=0

(60)(9.8)+T_L*3.5=0

T_L= -470.4N

Replacing in the first equation,

T_R+T_L=588N

T_R-470N=588N

T_R = 1058.4

Therefore the force at right side is 1058N

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RSB [31]

Answer:

r = 4.44 m

Explanation:

 

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Now let's use Newton's equilibrium relationship

         B - W = 0

         B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

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           A = 4π r²

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The volume of a sphere is

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Let's replace

     ρ g 4/3 π r³ = W₁ + σ 4π r²

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Let's replace the values

     r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

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As the independent term is very small we can despise it, to find the solution

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3 0
2 years ago
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Answer:

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Explanation:

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