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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 hoop of \(2.00-\mathrm{m}\) circumference is rolling down an inclined plane of length \(10.0 \mathrm{m}\) in a time of \(10.0 \mathrm{s} .\) It started out from rest. (a) What is its angular velocity when it arrives at the bottom? (b) If the mass of the hoop, concentrated at the rim, is \(1.50 \mathrm{kg},\) what is the angular momentum of the hoop when it reaches the bottom of the incline? (c) What force(s) supplied the net torque to change the hoop's angular momentum? Explain. [Hint: Use a rotation axis through the hoop's center. \(]\) (d) What is the magnitude of this force?
A stone used to grind wheat into flour is turned through 12 revolutions by a constant force of \(20.0 \mathrm{N}\) applied to the rim of a 10.0 -cm-radius shaft connected to the wheel. How much work is done on the stone during the 12 revolutions?
A sign is supported by a uniform horizontal boom of length \(3.00 \mathrm{m}\) and weight \(80.0 \mathrm{N} .\) A cable, inclined at an angle of \(35^{\circ}\) with the boom, is attached at a distance of \(2.38 \mathrm{m}\) from the hinge at the wall. The weight of the sign is \(120.0 \mathrm{N} .\) What is the tension in the cable and what are the horizontal and vertical forces \(F_{x}\) and \(F_{y}\) exerted on the boom by the hinge? Comment on the magnitude of \(F_{y}\).
Verify that the units of the rotational form of Newton's second law [Eq. (8-9)] are consistent. In other words, show that the product of a rotational inertia expressed in \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular acceleration expressed in \(\mathrm{rad} / \mathrm{s}^{2}\) is a torque expressed in \(\mathrm{N} \cdot \mathrm{m}\).
A 124 -g mass is placed on one pan of a balance, at a point \(25 \mathrm{cm}\) from the support of the balance. What is the magnitude of the torque about the support exerted by the mass?
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