/*! 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 17 A \(5000-\) lb truck is being us... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 17

A \(5000-\) lb truck is being used to lift a 1000 -lb boulder \(B\) that is on a 200 -lb pallet \(A\). Knowing the acceleration of the truck is \(1 \mathrm{ft} / \mathrm{s}^{2}\), determine \((a)\) the horizontal force between the tires and the ground, (b) the force between the boulder and the pallet.

Short Answer

Expert verified
(a) 192.55 lb, (b) 37.27 lb.

Step by step solution

01

Identify the Forces

First, identify all the forces acting on the system. The truck has a weight of 5000 lb, the boulder has a weight of 1000 lb, and the pallet has a weight of 200 lb. These weights act downward due to gravity. The truck also exerts a horizontal force to accelerate.
02

Calculate Total Mass

The total mass being accelerated by the truck includes the truck itself, the boulder, and the pallet. Convert these weights to masses using the relation: \( m = \frac{W}{g} \), where \( g = 32.2 \frac{ft}{s^2} \). Thus, the total mass \(M_{total}\) is:\[ M_{total} = \frac{5000 + 1000 + 200}{32.2} \approx 192.55 \text{ slugs} \]
03

Use Newton's Second Law for Horizontal Force

Apply Newton鈥檚 second law to find the horizontal force \( F \). The equation is given by: \( F = M_{total} \cdot a \), with \( a = 1 \frac{ft}{s^2} \). So:\[ F = 192.55 \times 1 = 192.55 \text{ lb} \]
04

Determine Force on Boulder by Pallet

To find the force between the boulder and the pallet, consider only the boulder and the pallet. The force required to accelerate them together horizontally is found using:\[ M_{boulder\, +\, pallet} = \frac{1000 + 200}{32.2} = 37.27 \text{ slugs} \]Thus, the force is:\[ F_{b+p} = 37.27 \times 1 = 37.27 \text{ lb} \]

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 91影视!

Key Concepts

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

Forces in Mechanics
Understanding forces in mechanics is crucial for analyzing problems involving motion. Forces are interactions that can change the motion of an object. They are vector quantities, meaning they have both magnitude and direction.

In mechanics, several types of forces can be acting on objects:
  • Gravitational force: This is the force pulling objects towards the center of the Earth, commonly perceived as weight.
  • Normal force: Acts perpendicular to the surface that an object is in contact with.
  • Frictional force: Resists the motion of an object sliding on a surface.
  • Applied force: Exerted by objects or entities, such as the truck's force in the problem.
When analyzing a system of forces, as with the truck, boulder, and pallet, it is important to identify all acting forces. This process often involves breaking down the forces into components and considering their impact in different directions.
Mass and Weight Conversion
In physics, understanding the difference between mass and weight is essential. Mass is the measure of the amount of matter in an object and does not change regardless of the object's location. The weight of an object, on the other hand, is the force exerted by gravity on it.

To convert weight to mass in the imperial system, use the formula: \[ m = \frac{W}{g} \] where \( W \) is the weight in pounds and \( g \) is the gravitational acceleration (32.2 ft/s虏).
  • For example, the truck in the problem has a weight of 5000 lb, which converts to mass by dividing by the gravitational constant.
  • This conversion is crucial for calculating how these objects behave under different forces, especially when using Newton's second law.
This distinction and conversion make analyzing forces and motions in problems much more manageable, ensuring accurate results.
Horizontal Force Calculation
Calculating horizontal forces often involves applying Newton's second law of motion, which relates net force, mass, and acceleration. According to Newton's second law, the formula is: \[ F = M \cdot a \] Where \( F \) is the net force applied, \( M \) is the mass, and \( a \) is the acceleration.

In the exercise, to determine the horizontal force exerted by the truck's wheels on the ground, you first calculate the total mass of the truck, boulder, and pallet.
  • Once the total mass is computed using weight-to-mass conversion, the acceleration given in the problem (1 ft/s虏) helps establish the horizontal force.
  • The calculated force of 192.55 lb represents the necessary force to move the entire system horizontally at the given acceleration.
  • Understanding these calculations helps in realizing how much force is required to move different mass combinations horizontally.
This approach is vital in engineering applications, helping to design systems capable of moving objects efficiently.

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 baggage conveyor is used to unload luggage from an airplane. The 10 -kg duffel bag \(A\) is sitting on top of the \(20-\) kg suitcase \(B .\) The conveyor is moving the bags down at a constant speed of \(0.5 \mathrm{m} / \mathrm{s}\) when the belt suddenly stops. Knowing that the cofficient of friction between the belt and \(B\) is 0.3 and that bag \(A\) does not slip on suitcase \(B\), determine the smallest allowable coefficient of static friction between the bags.

Pin \(B\) weighs 4 oz and is free to slide in a horizontal plane along the rotating arm \(O C\) and along the fixed circular slot \(D E\) of radius \(b=20\) in. Neglecting friction and assuming that \(\theta=15\) rad/s and \(\ddot{\theta}=250 \mathrm{rad} / \mathrm{s}^{2}\) for the position \(\theta=20^{\circ},\) determine for that position \((a)\) the radial and transverse components of the resultant force exerted on pin \(B,(b)\) the forces \(P\) and \(Q\) exerted on pin \(B\), respectively, by rod \(O C\) and the wall of slot \(D E .\)

A small ball swings in a horizontal circle at the end of a cord of length \(l_{1},\) which forms an angle \(\theta_{1}\) with the vertical. The cord is then slowly drawn through the support at \(O\) until the length of the free end is \(l_{2}\). (a) Derive a relation among \(l_{1}, l_{2}, \theta_{1},\) and \(\theta_{2} .\) (b) If the ball is set in motion so that initially \(l_{1}=0.8 \mathrm{m}\) and \(\theta_{1}=35^{\circ},\) determine the angle \(\theta_{2}\) when \(l_{2}=0.6 \mathrm{m} .\)

A space probe is to be placed in a circular orbit of radius \(4000 \mathrm{km}\) about the planet Mars. As the probe reaches \(A\), the point of its original trajectory closest to Mars, it is inserted into a first elliptic transfer orbit by reducing its speed. This orbit brings it to point \(B\) with a much- reduced velocity. There the probe is inserted into a second transfer orbit by further reducing its speed. Knowing that the mass of Mars is 0.1074 times the mass of the earth, that \(r_{A}=9000 \mathrm{km}\) and \(r_{B}=180000 \mathrm{km},\) and that the probe approaches \(A\) on a parabolic trajectory, determine the time needed for the space probe to travel from \(A\) to \(B\) on its first transfer orbit.

A 4 -kg projectile is fired vertically with an initial velocity of \(90 \mathrm{m} / \mathrm{s}\), reaches a maximum height, and falls to the ground. The aerodynamic drag \(\mathrm{D}\) has a magnitude \(D=0.0024 \mathrm{v}^{2}\) where \(D\) and \(v\) are expressed in newtons and \(\mathrm{m} / \mathrm{s}\), respectively. Knowing that the direction of the drag is always opposite to the direction of the velocity, determine \((a)\) the maximum height of the trajectory, (b) the speed of the projectile when it reaches the ground.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Thermodynamics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Physics of Motion

Read Explanation

Work Energy and Power

Read Explanation

Modern 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.