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Chapter 8: Problem 12

The pull cord of a lawnmower engine is wound around a drum of radius $6.00 \mathrm{cm} .\( While the cord is pulled with a force of \)75 \mathrm{N}$ to start the engine, what magnitude torque does the cord apply to the drum?

Short Answer

Expert verified
Answer: The magnitude of the torque applied to the drum by the pull cord is 4.50 N鈰卪.

Step by step solution

01

Identify the given information

We have the following information given in the problem: - Radius of the drum: \(r = 6.00\,\text{cm}\) - Force applied to the pull cord: \(F = 75\,\text{N}\)
02

Convert the radius to meters

To calculate the torque, we need the radius in meters. To convert the radius from centimeters to meters, we can use the following conversion: \(1 \, \mathrm{m} = 100\, \mathrm{cm}\) Therefore, the radius in meters is: \(r = 6.00\, \mathrm{cm} \times \frac{1 \, \mathrm{m}}{100 \, \mathrm{cm}} = 0.0600\,\mathrm{m}\)
03

Calculate the torque

Now, we can use the formula for torque, which is: \(\tau = r \times F\) Plugging in the values for the radius and the force, we get: \(\tau = (0.0600 \, \mathrm{m}) \times (75 \, \mathrm{N})\) Calculating the torque: \(\tau = 4.50\,\mathrm{N} \cdot \mathrm{m}\)
04

State the result

The magnitude of the torque applied to the drum by the pull cord is \(4.50\,\mathrm{N} \cdot \mathrm{m}\).

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