Question

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

Answer: The magnitude of the torque applied to the drum by the pull cord is 4.50 N⋅m.

Step-by-step solution

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: \(F = 75\,\text{N}\)

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: \(r = 6.00\, \mathrm{cm} \times \frac{1 \, \mathrm{m}}{100 \, \mathrm{cm}} = 0.0600\,\mathrm{m}\)

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: \(\tau = 4.50\,\mathrm{N} \cdot \mathrm{m}\)

State the result

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