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Chapter 8: Problem 11

A mechanic turns a wrench using a force of \(25 \mathrm{N}\) at a distance of \(16 \mathrm{cm}\) from the rotation axis. The force is perpendicular to the wrench handle. What magnitude torque does she apply to the wrench?

Short Answer

Expert verified
Answer: The magnitude of the torque applied to the wrench is 4 Nm.

Step by step solution

01

Identify the given information

We are given two important pieces of information: - Force applied on the wrench handle: \(F = 25 \mathrm{N}\) - Distance from the rotation axis to the point where the force is applied: \(d = 16 \mathrm{cm}\) (or \(0.16 \mathrm{m}\))
02

Recall the formula for torque

The formula to calculate the torque (\(\tau\)) is given by: $$\tau = F \cdot d \cdot \sin(\theta)$$ where \(\theta\) is the angle between the force and the lever arm. In our case, since the force is perpendicular to the wrench handle, the angle \(\theta = 90^{\circ}\) and the sine of \(\theta\) is equal to 1: $$\sin(90^{\circ}) = 1$$
03

Calculate the torque

Now that we have all necessary information and the formula, we can calculate the magnitude of the torque as follows: $$\tau = F \cdot d \cdot \sin(\theta) = 25 \mathrm{N} \cdot 0.16 \mathrm{m} \cdot 1 = 4 \mathrm{Nm}$$ So, the magnitude of the torque applied to the wrench is \(4 \mathrm{Nm}\).

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Most popular questions from this chapter

A hollow cylinder, of radius \(R\) and mass \(M,\) rolls without slipping down a loop-the-loop track of radius \(r .\) The cylinder starts from rest at a height \(h\) above the horizontal section of track. What is the minimum value of \(h\) so that the cylinder remains on the track all the way around the loop? A hollow cylinder, of radius \(R\) and mass \(M,\) rolls without slipping down a loop-the- loop track of radius \(r .\) The cylinder starts from rest at a height \(h\) above the horizontal section of track. What is the minimum value of \(h\) so that the cylinder remains on the track all the way around the loop?
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