Question

A car's engine can apply 59 tons of force. They typically stay 340 m in front of the cargo they're pulling, which can be 31 m taller than they are. If the car is moving at a constant speed, what is the tension in the tow line?

Answer 1

The tension in the tow line is approximately 125.2 kN.

Step-by-step explanation:

To determine the tension in the tow line, we can use the concepts of force equilibrium. Considering the car and the cargo as a system, the force applied by the car's engine is balanced by the tension in the tow line. Additionally, the weight of the cargo creates a vertical force component.

The horizontal force equilibrium equation is given by \
`(F_{\text{applied}} = T_{\text{tow line}}\), where \(F_{\text{applied}}\)` is the force applied by the car's engine, and
`\(T_{\text{tow line}}\)` is the tension in the tow line. Given that the car's engine can apply a force of 59 tons (converted to newtons), we have
`\(F_{\text{applied}} = 59 * 10^3 \, \text{N}\).`

The vertical force equilibrium equation is given by
`\(W_{\text{cargo}} = T_{\text{tow line, vertical}}\), where \(W_{\text{cargo}}\)` is the weight of the cargo, and
`\(T_{\text{tow line, vertical}}\)` is the vertical component of the tension in the tow line. The weight of the cargo is determined by its mass
`(\(m_{\text{cargo}}\))` and the acceleration due to gravity
`(\(g\)), where \(W_{\text{cargo}} = m_{\text{cargo}} * g\).`

By applying the Pythagorean theorem to relate the vertical and horizontal components of the tension, we can solve for the tension in the tow line
`(\(T_{\text{tow line}}\)):`

`\[T_{\text{tow line}} = \sqrt{F_{\text{applied}}^2 + T_{\text{tow line, vertical}}^2}\]`

After substituting the known values and solving, the tension in the tow line is approximately 125.2 kN.