/*! 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 block with a weight of \(3.0 \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A block with a weight of \(3.0 \mathrm{~N}\) is at rest on a horizontal surface. A \(1.0 \mathrm{~N}\) upward force is applied to the block by means of an attached vertical string. What are the (a) magnitude and (b) direction of the force of the block on the horizontal surface?

Short Answer

Expert verified
The force of the block on the horizontal surface is 2.0 N downward.

Step by step solution

01

Understanding the Problem

First, identify the forces acting on the block. There is a gravitational force acting downward with a magnitude of \(3.0 \, \text{N}\) and an applied upward force of \(1.0 \, \text{N}\). The block is resting on a horizontal surface, so the normal force from the surface acts upward as well.
02

Determine the Net Force

Since the block is at rest, the net force acting on it is zero. Therefore, the sum of the forces in the vertical direction must be zero.
03

Calculate Forces on the Block

The gravitational force \( F_g = 3.0 \, \text{N} \) acts downward, and the upward string force \( F_{up} = 1.0 \, \text{N} \) acts upward. The normal force \( F_n \) from the surface will balance out these forces.
04

Equation Setup

Set up the equation for the vertical forces: \( F_n + F_{up} = F_g \), substituting the known values: \( F_n + 1.0 \, \text{N} = 3.0 \, \text{N} \).
05

Solve for the Normal Force

Solve the equation to find \( F_n \): \( F_n = 3.0 \, \text{N} - 1.0 \, \text{N} = 2.0 \, \text{N} \).
06

Determine the Direction of the Force

Since \( F_n \) is the force exerted by the surface to support the block, it is directed upward, opposite to the gravitational force.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Newton's Laws of Motion
Newton's laws of motion are foundational principles in classical mechanics, describing the relationship between a body and the forces acting upon it. These laws are key to understanding how objects move and interact.
Newton's first law, also known as the law of inertia, states that an object at rest stays at rest, and an object in motion stays in motion at the same speed and in the same direction unless acted upon by an unbalanced external force. This principle helps us understand why the block remains stationary in the given exercise if the forces are balanced.
Newton's second law provides the formula to calculate the net force acting on an object. It states that the acceleration of an object depends on the net force acting upon it and its mass, expressed as: \[ F = ma \]In the exercise, since the block is not accelerating, the net force is zero, indicating that the upward and downward forces are balanced.
Newton's third law states that for every action, there is an equal and opposite reaction. This law is exemplified by the normal force exerted by the surface on the block, which balances the gravitational force and the applied upward force.
Forces and Motion
In physics, forces are vectors, which means they have both magnitude and direction. These two properties play a crucial role in determining an object's motion.
In the given exercise, several forces act on the block, including:
  • Gravitational force: Pulls the block downward towards the center of the earth.
  • Upward force from the string: Pulls the block upwards in opposition to gravity.
  • Normal force from the surface: Acts perpendicularly to the surface, balancing the other forces.
The block in the exercise remains at rest due to the equilibrium of these forces. The total forces acting upwards match the total forces acting downwards, maintaining the block's stationary state.
The understanding of forces and motion helps predict how objects behave when forces act upon them, allowing accurate calculations in designing structures or in everyday situations.
Equilibrium of Forces
Equilibrium occurs when all the forces acting on an object are perfectly balanced, resulting in no net force and no change in motion. This happens when the sum of forces in any direction equals zero.
In the exercise, the block is in equilibrium on the horizontal surface. The sum of all forces in the vertical direction is zero:\[ F_{net} = F_n + F_{up} - F_g = 0 \]Substituting the known forces:\[ F_n = F_g - F_{up} \]By calculating: \( F_n = 3.0\, \text{N} - 1.0\, \text{N} = 2.0\, \text{N} \), we find that the normal force is precisely the amount needed to balance those forces acting downward.
Recognizing equilibrium situations is vital in structural engineering and physics, ensuring systems are stable and secure. It provides insight into how forces interact in real-world scenarios, allowing us to predict and control motion and stability.

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 spaceship lifts off vertically from the Moon, where \(g=\) \(1.6 \mathrm{~m} / \mathrm{s}^{2}\). If the ship has an upward acceleration of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) as it lifts off, what is the magnitude of the force exerted by the ship on its pilot, who weighs \(735 \mathrm{~N}\) on Earth?

Tarzan, who weighs \(820 \mathrm{~N}\), swings from a cliff at the end of a \(20.0 \mathrm{~m}\) vine that hangs from a high tree limb and initially makes an angle of \(22.0^{\circ}\) with the vertical. Assume that an \(x\) axis extends horizontally away from the cliff edge and a \(y\) axis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is \(760 \mathrm{~N}\). Just then, what are (a) the force on him from the vine in unit- vector notation and the net force on him (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis? What are the (e) magnitude and (f) angle of Tarzan's acceleration just then?

While two forces act on it, a particle is to move at the constant velocity \(\vec{v}=(3 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(4 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). One of the forces is \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((-6 \mathrm{~N}) \hat{\mathrm{j}} .\) What is the other force?

A block is projected up a frictionless inclined plane with initial speed \(v_{0}=3.50\) plane with initial speed \(v_{0}=3.50\) \(\mathrm{m} / \mathrm{s}\). The angle of incline is \(\theta=32.0^{\circ} .\) (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m}\). He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)

See all solutions

Recommended explanations on Physics Textbooks

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.