/*! 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 88 The path of motion of a 5-lb par... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 88

The path of motion of a 5-lb particle in the horizontal plane is described in terms of polar coordinates as \(r=(2 t+1) \mathrm{ft}\) and \(\theta=\left(0.5 t^{2}-t\right) \mathrm{rad}\), where \(t\) is in seconds. Determine the magnitude of the unbalanced force acting on the particle when \(t=2 \mathrm{~s}\).

Short Answer

Expert verified
The magnitude of the unbalanced force acting on the particle when \(t=2\) s is 1.6 lbs.

Step by step solution

01

Determine Radial and Angular Components

Start by substituting \(t = 2\) s into \(r = 2t + 1\) and \(\theta = 0.5t^2 - t\). This gives: \( r = 2(2) + 1 = 5 ft\) and \(\theta = 0.5(2)^2 - 2 = 0 rad\). Also, find the first derivative of \(r\) and \(\theta\) to obtain their velocities: \( r' = 2 ft/s\) and \( \theta' = t - 1\) s. Substituting \(t = 2\) s into \(\theta'\) gives: \( \theta' = 2 - 1 = 1 rad/s\). Additionally, find the second derivative of \(r\) and \(\theta\) to determine their accelerations. We find that \(r'' = 0\) and \(\theta'' = 1\).
02

Compute Radial and Transverse Components of Acceleration

The radial and transverse acceleration in polar coordinates are given by: \(a_r = r'' - r(\theta')^2\) and \(a_\theta = r\theta'' + 2r'\theta'\). Substituting the previously obtained values gives: \(a_r = 0 - 5(1)^2 = -5 ft/s^2\) and \(a_\theta = 5(1) + 2*2*1 = 9 ft/s^2\).
03

Determine the Unbalanced Force

According to Newton's second law \( F = m a \), where \( F \) is the net force, \( m \) is the mass and \( a \) is the acceleration. Since the mass is given as 5lbs, we need to convert this to slug using the relation 1 lbs = 0.031 slug, so m = 5 lbs * 0.031 = 0.155 slug. The total acceleration will be: \( a = \sqrt{a_r^2 + a_\theta^2} = \sqrt{(-5)^2 + 9^2} = 10.3 ft/s^2 \). Using this, the unbalanced force is: \( F = m * a = 0.155 slug * 10.3 ft/s^2 = 1.6 lbs\).

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!

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

The 150-lb man lies against the cushion for which the coefficient of static friction is \(\mu_{s}=0.5 .\) If he rotates about the \(z\) axis with a constant speed \(v=30 \mathrm{ft} / \mathrm{s}\) determine the smallest angle \(\theta\) of the cushion at which he will begin to slip off. Probs. \(13-71 / 72\)

The 150-lb man lies against the cushion for which the coefficient of static friction is \(\mu_{s}=0.5 .\) Determine the resultant normal and frictional forces the cushion exerts on him if, it due to rotation about the \(z\) axis, he has a constant speed \(v=20 \mathrm{ft} / \mathrm{s}\). Neglect the size of the man. Take \(\theta=60^{\circ}\).

The conveyor belt delivers each \(12-\mathrm{kg}\) crate to the ramp at \(A\) such that the crate's speed is \(v_{A}=2.5 \mathrm{~m} / \mathrm{s}\), directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is \(\mu_{k}=0.3\), determine the smallest incline \(\theta\) of the ramp so that the crates will slide off and fall into the cart. Probs. \(13-20 / 21\)

The \(300-\mathrm{kg}\) bar \(B\), originally at rest, is being towed over a series of small rollers. Determine the force in the cable when \(t=5 \mathrm{~s}\), if the motor \(M\) is drawing in the cable for a short time at a rate of \(v=\left(0.4 t^{2}\right) \mathrm{m} / \mathrm{s}\), where \(t\) is in seconds \((0 \leq t \leq 6 \mathrm{~s})\). How far does the bar move in \(5 \mathrm{~s}\) ? Neglect the mass of the cable, pulley, and the rollers. Prob. 13-38

The forked rod is used to move the smooth 2 -lb particle around the horizontal path in the shape of a lima莽on, \(r=(2+\cos \theta) \mathrm{ft}\). If \(\theta=\left(0.5 t^{2}\right) \mathrm{rad}\), where \(t\) is in seconds, determine the force which the rod exerts on the particle at the instant \(t=1 \mathrm{~s}\). The fork and path contact the particle on only one side. Prob. 13-107
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Nuclear Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Torque and Rotational Motion

Read Explanation

Measurements

Read Explanation

Translational Dynamics

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.