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FinnZ [79.3K]
3 years ago
15

A square metal block with mass (m) is suspended by five wires with the same length. One of these wires is made of iron at the ce

nter of the block and has a radius (1.2 mm) while the other four wires have the same cross-sectional area and they are made of copper at the block corners. If the tension in the iron wire equals (0.3 mg), find the radius of a copper wire. (Take Yiron = 2×1011, Ycopper =1. 2×1011 N/m2 )
Physics
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

1.18 mm

Explanation:

The mass of the block is m, so the weight of the block is mg.  If the tension in the iron wire is 0.3 mg, then the total tension in the 4 copper wires is 0.7 mg.  So the tension in each copper wire is 0.175 mg.

For each wire:

ΔL/L = F / (EA)

where ΔL is the change in length,

L is the initial length,

F is the force,

E is Young's modulus (modulus of elasticity),

and A is the cross sectional area.

The five wires have the same initial length, and stretch by the same amount.

ΔL/L = ΔL/L

F / (EA) = F / (EA)

(0.3 mg) / (2×10¹¹ Pa × π(1.2 mm)²) = (0.175 mg) / (1.2×10¹¹ Pa × πr²)

0.3 / (2 (1.2 mm)²) = 0.175 / (1.2 r²)

0.36 r² = 0.504 mm²

r = 1.18 mm

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Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

\Sigma F = F - F' = M\cdot a

Box with mass 2M

\Sigma F = F' - F'' = 2\cdot M \cdot a

Box with mass 3M

\Sigma F = F'' = 3\cdot M \cdot a

On the third equation, acceleration can be modelled in terms of F'':

a = \frac{F''}{3\cdot M}

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

F' = 2\cdot M \cdot a + F''

F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''

F' = \frac{5}{3}\cdot F''

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)

F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''

F = 2\cdot F''

F'' = \frac{1}{2}\cdot F

Afterwards, F' as function of the external force can be obtained by direct substitution:

F' = \frac{5}{6}\cdot F

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