Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 33

A woman who weighs \(600 \mathrm{N}\) sits on a chair with her feet on the floor and her arms resting on the chair's armrests. The chair weighs $100 \mathrm{N}\(. Each armrest exerts an upward force of \)25 \mathrm{N}$ on her arms. The seat of the chair exerts an upward force of \(500 \mathrm{N}\). (a) What force does the floor exert on her feet? (b) What force does the floor exert on the chair? (c) Consider the woman and the chair to be a single system. Draw an FBD for this system that includes all of the external forces acting on it.

Short Answer

Expert verified
Answer: The force exerted by the floor on the woman's feet is 50 N and the force exerted by the floor on the chair is 100 N.

Step by step solution

01

Calculate the total upward force exerted by the chair on the woman

To find the total upward force exerted by the chair on the woman, simply add the forces exerted by the seat and armrests of the chair. Thus: Total Upward Force = Force_seat + 2 * Force_armrests = \(500 N + 2 * 25 N\) = \(500 N + 50 N\) = \(550 N\)
02

Find the force exerted by the floor on the woman's feet (a)

Since the woman is in equilibrium (not accelerating), the total upward force acting on her should be equal to the total downward force exerted by her weight. The force exerted by the floor on her feet can be calculated by subtracting the total upward force from her weight: Force_floor_woman = Weight_woman - Total Upward Force = \(600 N - 550 N\) = \(50 N\) The force exerted by the floor on the woman's feet is \(50 N\).
03

Find the force exerted by the floor on the chair (b)

As the chair is also in equilibrium (not accelerating), the downward force (weight of the chair) should be equal to the upward force exerted by the floor on the chair: Force_floor_chair = Weight_chair = \(100 N\) The force exerted by the floor on the chair is \(100 N\).
04

Draw a Free Body Diagram for the combined system (c)

In order to draw an accurate Free Body Diagram (FBD) for the combined system, we consider the external forces acting on the woman and chair system. These forces include: 1. The force exerted by the floor on the woman's feet (upward, \(50 N\)) 2. The force exerted by the floor on the chair (upward, \(100 N\)) 3. The weight of the woman (downward, \(600 N\)) 4. The weight of the chair (downward, \(100 N\)) Label these forces appropriately, with vectors pointing in the respective directions (upward and downward). Drawing arrows to indicate these forces, you can create a Free Body Diagram representing the external forces acting on this combined system.

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 skydiver, who weighs \(650 \mathrm{N},\) is falling at a constant speed with his parachute open. Consider the apparatus that connects the parachute to the skydiver to be part of the parachute. The parachute pulls upward with a force of \(620 \mathrm{N} .\) (a) What is the force of the air resistance acting on the skydiver? (b) Identify the forces and the interaction partners of each force exerted on the skydiver. (c) Identify the forces and interaction partners of each force exerted on the parachute.
A horse is trotting along pulling a sleigh through the snow. To move the sleigh, of mass \(m,\) straight ahead at a constant speed, the horse must pull with a force of magnitude \(T.\) (a) What is the net force acting on the sleigh? (b) What is the coefficient of kinetic friction between the sleigh and the snow?
A 6.0 -kg block, starting from rest, slides down a frictionless incline of length \(2.0 \mathrm{m} .\) When it arrives at the bottom of the incline, its speed is \(v_{\mathrm{f}} .\) At what distance from the top of the incline is the speed of the block \(0.50 v_{\mathrm{f}} ?\)
A man lifts a 2.0 -kg stone vertically with his hand at a constant upward velocity of \(1.5 \mathrm{m} / \mathrm{s} .\) What is the magnitude of the total force of the man's hand on the stone?
A truck driving on a level highway is acted on by the following forces: a downward gravitational force of \(52 \mathrm{kN}\) (kilonewtons); an upward contact force due to the road of \(52 \mathrm{kN} ;\) another contact force due to the road of \(7 \mathrm{kN},\) directed east; and a drag force due to air resistance of \(5 \mathrm{kN}\), directed west. What is the net force acting on the truck?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Atoms and Radioactivity

Read Explanation

Modern Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Thermodynamics

Read Explanation

Physics of Motion

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