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Chapter 8: Problem 15

A uniform door weighs \(50.0 \mathrm{N}\) and is \(1.0 \mathrm{m}\) wide and 2.6 m high. What is the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner?

Short Answer

Expert verified
Answer: The magnitude of the torque is 69.5 Nm.

Step by step solution

01

Identify the forces acting on the door

The only force acting on the door is its weight. Since the door is uniform, the weight of the door acts at its center of mass. The center of mass is at the midpoint of the door, i.e., half the width and half the height of the door.
02

Calculate the location of center of mass

The center of mass can be calculated by dividing both width and height by 2. Therefore, the center of mass coordinates for this door are (0.5m,1.3m).
03

Calculate the lever arm

The lever arm is the perpendicular distance between the axis of rotation (one of the corners of the door) and the line of action of force, which is the weight of the door acting through the center of mass. To find this distance, we can use the Pythagorean theorem. The lever arm, \(r\), can be calculated as \(r = \sqrt{(0.5m)^2 + (1.3m)^2} = \sqrt{0.25 + 1.69} = \sqrt{1.94}\)
04

Calculate the torque

Now, we can find the torque τ, by using the torque formula, τ = Force × lever arm (perpendicular distance). In this case, the force is the weight of the door which is 50N. So, the torque τ can be calculated as: \(\tau = (50.0 \ \mathrm{N}) \times \sqrt{1.94 \ \mathrm{m^2}}\)
05

Solve for the magnitude of the torque

Now, we can find the magnitude of the torque due to the door's own weight by calculating the product of the force and the lever arm: \(\tau = 50.0 \ \mathrm{N} \times \sqrt{1.94 \ \mathrm{m^2}} = 50.0 \ \mathrm{N} \times 1.39 \ \mathrm{m} = 69.5 \ \mathrm{Nm}\) Therefore, the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner is 69.5 Nm.

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Most popular questions from this chapter

The radius of a wheel is \(0.500 \mathrm{m} .\) A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude $5.00 \mathrm{N},$ unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope. (a) How much rope unwinds while the wheel makes 1.00 revolution? (b) How much work is done by the rope on the wheel during this time? (c) What is the torque on the wheel due to the rope? (d) What is the angular displacement \(\Delta \theta\), in radians, of the wheel during 1.00 revolution? (e) Show that the numerical value of the work done is equal to the product \(\tau \Delta \theta\)
A 46.4 -N force is applied to the outer edge of a door of width $1.26 \mathrm{m}$ in such a way that it acts (a) perpendicular to the door, (b) at an angle of \(43.0^{\circ}\) with respect to the door surface, (c) so that the line of action of the force passes through the axis of the door hinges. Find the torque for these three cases.
A turntable must spin at 33.3 rpm \((3.49 \mathrm{rad} / \mathrm{s})\) to play an old-fashioned vinyl record. How much torque must the motor deliver if the turntable is to reach its final angular speed in 2.0 revolutions, starting from rest? The turntable is a uniform disk of diameter \(30.5 \mathrm{cm}\) and mass \(0.22 \mathrm{kg}\).
Two children standing on opposite sides of a merry-goround (see Fig. 8.5 ) are trying to rotate it. They each push in opposite directions with forces of magnitude \(10.0 \mathrm{N} .\) (a) If the merry-go-round has a mass of $180 \mathrm{kg}\( and a radius of \)2.0 \mathrm{m},$ what is the angular acceleration of the merry-go-round? (Assume the merry-go-round is a uniform disk.) (b) How fast is the merry-go-round rotating after \(4.0 \mathrm{s} ?\)
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?
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