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Chapter 4: Problem 96

A helicopter is lifting two crates simultaneously. One crate with a mass of \(200 \mathrm{kg}\) is attached to the helicopter by a cable. The second crate with a mass of \(100 \mathrm{kg}\) is hanging below the first crate and attached to the first crate by a cable. As the helicopter accelerates upward at a rate of \(1.0 \mathrm{m} / \mathrm{s}^{2},\) what is the tension in each of the two cables?

Short Answer

Expert verified
The tension in the first cable, attaching the first crate to the helicopter, is 3040 N. The tension in the second cable, attaching the second crate to the first crate, is 880 N.

Step by step solution

01

Draw a free-body diagram for both crates

Draw a free-body diagram for each crate. Include the weight of each crate (due to gravity), the tension in the cables, and the net force acting on them due to the upward acceleration.
02

Write force equations for Newton's second law for each crate

According to Newton's second law, the net force on an object is equal to the product of its mass and acceleration \((F=ma)\). Use this law to create force equations for each crate. For the first crate: \(\Sigma F_{1} = T_{1} - T_{2} - W_{1} = m_{1}a\) For the second crate: \(\Sigma F_{2} = -T_{2} - W_{2} = m_{2}a\) where \(T_{1}\) is the tension in the cable attaching the first crate to the helicopter, \(T_{2}\) is the tension in the cable attaching the second crate to the first crate, \(W_{1}\) and \(W_{2}\) are the weights of the first and second crates due to gravity, respectively, \(m_{1}\) and \(m_{2}\) are the masses of the first and second crates, respectively, and \(a\) is the upward acceleration.
03

Solve for the weights of both crates

The weights of the crates can be calculated using the formula: \(W = mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity \((9.8\, \mathrm{m/s^2}).\) For the first crate: \(W_{1} = m_{1}g = (200\, \mathrm{kg})(9.8\, \mathrm{m/s^2}) = 1960\, \mathrm{N}\) For the second crate: \(W_{2} = m_{2}g = (100\, \mathrm{kg})(9.8\, \mathrm{m/s^2}) = 980\, \mathrm{N}\)
04

Solve for the tension in the second cable

Use the force equation for the second crate (\(\Sigma F_{2} = -T_{2} - W_{2} = m_{2}a\)) to solve for the tension in the cable attaching the second crate to the first crate (\(T_{2}\)). \(-T_{2} - 980\, \mathrm{N} = (100\, \mathrm{kg})(1.0\, \mathrm{m/s^2})\) \(-T_{2} - 980\, \mathrm{N} = 100\, \mathrm{N}\) \(T_{2} = 980\, \mathrm{N} - 100\, \mathrm{N} = 880\, \mathrm{N}\)
05

Solve for the tension in the first cable

Use the force equation for the first crate (\(\Sigma F_{1} = T_{1} - T_{2} - W_{1} = m_{1}a\)) and the value of the tension found in step 4 to solve for the tension in the cable attaching the first crate to the helicopter (\(T_{1}\)). \(T_{1} - 880\, \mathrm{N} - 1960\, \mathrm{N} = (200\, \mathrm{kg})(1.0\, \mathrm{m/s^2})\) \(T_{1} - 2840\, \mathrm{N} = 200\, \mathrm{N}\) \(T_{1} = 200\, \mathrm{N} + 2840\, \mathrm{N} = 3040\, \mathrm{N}\) The tension in the first cable (attaching the first crate to the helicopter) is \(3040\, \mathrm{N}\), and the tension in the second cable (attaching the second crate to the first crate) is \(880\, \mathrm{N}\).

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