Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 27

A rod is being used as a lever as shown. The fulcrum is \(1.2 \mathrm{m}\) from the load and \(2.4 \mathrm{m}\) from the applied force. If the load has a mass of \(20.0 \mathrm{kg}\). what force must be applied to lift the load?

Short Answer

Expert verified
Solution: To lift the load, a force of 98.1 N must be applied to the rod.

Step by step solution

01

Convert mass to weight

First, we need to convert the mass of the load (\(20.0 \mathrm{kg}\)) to weight, which is the force acting on the load due to gravity, by multiplying the mass by the acceleration due to gravity (\(g\approx 9.81 \mathrm{m/s^2}\)). Weight of the load \(= m \times g = 20.0 \mathrm{kg} \times 9.81 \mathrm{m/s^2} = 196.2 \mathrm{N}\)
02

Calculate the torque due to the load

We can now calculate the torque due to the load as the product of the weight of the load and the distance from the fulcrum: \(\tau_\text{load} = \text{Weight of the load} \times \text{Distance from the fulcrum} = 196.2 \mathrm{N} \times 1.2 \mathrm{m} = 235.44 \mathrm{Nm}\)
03

Find the applied force

In order to lift the load, the applied force must generate a torque equal in magnitude to that of the load but with an opposite direction. Since we know the distance of the applied force from the fulcrum (\(2.4 \mathrm{m}\)), we can calculate the required force: \(\tau_\text{applied} = \tau_\text{load} = F_\text{applied} \times \text{Distance from the fulcrum}\) \(F_\text{applied} = \frac{\tau_\text{load}}{\text{Distance from the fulcrum}} = \frac{235.44 \mathrm{Nm}}{2.4 \mathrm{m}} = 98.1 \mathrm{N}\) So, to lift the load, a force of \(98.1 \mathrm{N}\) must be applied to the rod.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 2.0 -kg uniform flat disk is thrown into the air with a linear speed of \(10.0 \mathrm{m} / \mathrm{s} .\) As it travels, the disk spins at 3.0 rev/s. If the radius of the disk is \(10.0 \mathrm{cm},\) what is the magnitude of its angular momentum?
A Ferris wheel rotates because a motor exerts a torque on the wheel. The radius of the London Eye, a huge observation wheel on the banks of the Thames, is \(67.5 \mathrm{m}\) and its mass is \(1.90 \times 10^{6} \mathrm{kg} .\) The cruising angular speed of the wheel is $3.50 \times 10^{-3} \mathrm{rad} / \mathrm{s} .$ (a) How much work does the motor need to do to bring the stationary wheel up to cruising speed? [Hint: Treat the wheel as a hoop.] (b) What is the torque (assumed constant) the motor needs to provide to the wheel if it takes 20.0 s to reach the cruising angular speed?
The angular momentum of a spinning wheel is $240 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ After application of a constant braking torque for \(2.5 \mathrm{s}\), it slows and has a new angular momentum of $115 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ What is the torque applied?
A 1.10 -kg bucket is tied to a rope that is wrapped around a pole mounted horizontally on friction-less bearings. The cylindrical pole has a diameter of \(0.340 \mathrm{m}\) and a mass of \(2.60 \mathrm{kg} .\) When the bucket is released from rest, how long will it take to fall to the bottom of the well, a distance of \(17.0 \mathrm{m} ?\)
A bicycle has wheels of radius \(0.32 \mathrm{m}\). Each wheel has a rotational inertia of \(0.080 \mathrm{kg} \cdot \mathrm{m}^{2}\) about its axle. The total mass of the bicycle including the wheels and the rider is 79 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Engineering Physics

Read Explanation

Oscillations

Read Explanation

Magnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Kinematics Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free