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A 82.0 kg worker climbs a cell tower which is 72.0 m tall. The worker brings...

A 82.0 kg worker climbs a cell tower which is 72.0 m tall. The worker brings 208 kg of tools and equipment to the top off the tower. The tower can be modeled as equivalent to a steel cylinder that has a cross sectional area of .022 m^2. By how much did the tower compress when the worker was at the top of the tower. Young's modulus for steel is 20.0 x 10^10 Pa.

A. .0465 mm

B. .0242 mm

C. .0345 mm

D. .0121 mm

Solutions

Expert Solution

Let the height of tower be represented as l and the height by which the tower is compressed is

Also,

and where = F is the force on the tower and A is the area of cross section of cylinder.

Also,

Here, Y = Young's modulus

So, putting the values of Stress and Strain, we get

Reorganising,

So,

Please refer to figure attached.

Here F is the net force on tower which is total mass of man and tool and gravity where is m is total mass of man and tool and g=9.8

Putting the values, we get

which is [Since so we multiply by to convert into mm ]

Hence, option A is correct.

genius_generous answered 2 months ago

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