/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 A person pushes on a door and it... [FREE SOLUTION] | 91Ó°ÊÓ

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A person pushes on a door and it swings open. Where should the force be applied in order to make the door swing open as quickly as possible? A. On the edge of the door nearest the hinges. B. At the center of the door. C. Oe the edge farthest from the hinges. D. A force anywhere on the door will have the same effect.

Short Answer

Expert verified
Apply the force on the edge farthest from the hinges.

Step by step solution

01

- Understanding Torque

Torque is a measure of the rotational force. It depends on two factors: the magnitude of the force applied and the distance from the pivot point (the hinges, in this case) to where the force is applied. Mathematically, it is given as \(\tau = r \times F\), where \(r\) is the distance from the pivot and \(F\) is the force.
02

- Analyzing Different Application Points

If the force is applied closest to the hinges, \(r\) is smallest, reducing the torque. If applied at the center, \(r\) is larger than the distance to the nearest edge but smaller than the farthest edge, providing moderate torque. Applying the force at the edge farthest from the hinges provides the largest \(r\), producing the maximum torque.
03

- Conclusion

To make the door swing open as quickly as possible, the force should be maximized. This means applying the force where \(r\) is largest, i.e., on the edge farthest from the hinges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Force
Rotational force, or torque, is what causes objects to rotate around a pivot point. Imagine pushing on a door to make it swing open. Your push is like a rotational force. If you push harder, the door swings faster, but where you push also matters.
The formula for torque is \(\tau = r \times F \) where \(\tau \) is torque, \(r \) is the distance from the axis of rotation (the pivot point), and \(F \) is the force applied. This means that both the amount of force and the distance from the pivot are important for generating torque.
Your push creates a force that travels to the pivot point, causing rotation. If you push far from the pivot, you get more torque. This is why it's easier to open a door when pushing at the edge rather than near the hinges.
Pivot Point
A pivot point is the spot around which everything rotates. When opening a door, the hinges act as the pivot point. They don't move, but everything else does.
Consider a door as a lever. The hinge is the fulcrum or pivot point. When you push on the door, the force you apply travels through the door to the hinges. This makes the door rotate around the hinges, making them the key part in the door's rotation.
Think about a seesaw. The middle point is its pivot. The farther you sit from it, the easier it is to lift the other end. The same idea applies to the door: the farther you push from the hinge (pivot), the easier it is to make the door swing open.
Distance from Axis of Rotation
The distance from the axis of rotation plays a major role in determining the force's effect on rotation. This distance is crucial in calculating torque.
If you apply the same force at different distances from the pivot point, the effect on rotation changes. The farther from the pivot you apply the force, the more torque you generate. This is why pushing a door near the edge is much more effective than pushing it near the hinges.
In our example, to make the door swing open quickly, you should apply the force at the point farthest from the hinges. This maximizes \(r \) in the formula \(\tau = r \times F \), thereby maximizing torque and causing the door to open faster and with less effort.

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Most popular questions from this chapter

Energy consumption in the home is generally measured in units of kilowatt hours. A kilowatt hour is equal to: A. \(3,600 \mathrm{~J}\) B. \(6,000 \mathrm{~J}\) C. \(3.600 .000 \mathrm{~J}\) D. \(6,000,000 \mathrm{~J}\)

A child pashes a block across the floor with a constant force of \(5 \mathrm{~N}\). The block moves in a straight line and its speed increases from \(0.2 \mathrm{~m} / \mathrm{s}\) to \(0.6 \mathrm{~m} / \mathrm{s}\). Which of the following must be true? A. The force applied by the child is greater than the force of kinetic friction between the block and the floct. B. The force applied by the child is less than the force of kinetic friction between the block and the floor. C. The force applied by the child is greater than the force due to the weight of the block. D. The force applied by the child is less than the force due to the weight of the block.

There are 3 forces acting on an object. Two of the forces are of equal magnitude. One of these forces pulls the object to the north and one pulls to the east. If the object undergoes no acceleration, then in which direction must the third force be pulling? A. northeast B. northwest C. southeast D. southwest

A carpenter who is having a difficult time loosening a serew puts away his screwdriver and chooses another with a bandle with a larger diameter. He does this because: A. increasing force increases torque- B. decreasing force decreases torque. C. increasing lever arm increases torque. D. decreasing lever am decreases torque.

A sign hangs by a rope attached at \(30^{\circ}\) to the middtle of its upper odge. It rests against a frictionless wall. If the weight of the sign were doubled, what would bappen to the tension in the string? (Note: \(\sin 30^{\circ}=0.5 ; \cos 30^{\circ}=0.87\) ) A. It would remain the same. B. It would increase by a factor of 1.5. C. It would increase by a factor of 2 . D. It would increase by a factor of 4 .

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