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Harman [31]
3 years ago
8

A 4.40-m-long, 500 kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new

building under construction. A 70 kg construction worker stands at the far end of the beam.
What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?
Physics
1 answer:
icang [17]3 years ago
4 0

Answer:

Torque=13798.4 N.m

Explanation:

Given data

Mass of beam m₁=500 kg

Mass of the person m₂=70 kg

length of steel r₁=4.40m

center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m

To find

Torque

Solution

Torque due to beam own weight

T_{1}=m_{1}gr_{1}\\ T_{1}=500*2.2*9.8\\T_{1}=10780N.m

Torque due to person

T_{2}=m_{2}r_{2}g\\T_{2}=70*(4.40)*(9.8)\\T_{2}=3018.4 N.m

Now for total torque

T_{total}=T_{1}+T_{2}\\T_{total}=10780+3018.4\\T_{total}=13798.4N.m

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Answer:

Part A:

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Part B:

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Part C:

\frac{\triangle m}{m}=0.004998=0.49985\%

Explanation:

Given Data:

v=20m/s

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Part A:

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Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

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Energy Required to move=200*43200=8640000 Joules

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K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

Kinetic Energy=\frac{1}{2}mv^2

Now, the relation between energies ratio and masses is:

\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2

\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977

\frac{\triangle m}{m}=0.004998=0.49985\%

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When he starts, let h = 0 ⇒ E₁ = 1/2mv₁²
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Without friction, energy is conserved at all times.

E₁ = E₂
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1/2mv₁² = mgh + 1/2mv₂²
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     ↓
gh = 1/2(v₁² - v₂²)
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h = (v₁² - v₂²) / (2g)


5 0
3 years ago
Read 2 more answers
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