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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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