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Q3.4-20E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 90

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

An object at rest on an inclined plane will not slide until the component of the gravitational force down the incline is sufficient to overcome the force due to static friction. Static friction is governed by an experimental law somewhat like that of kinetic friction (Problem 18); it has a magnitude of at most N, where m is the coefficient of static friction and N is, again, the magnitude of the normal force exerted by the surface on the object. If the plane is inclined at an angle a, determine the critical value ₀for which the object will slide if $a > a$ _obut will not move for $a < a$ _o.

Therefore, the object starts sliding down when ${tanα}_{o} > μ$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Draw a diagram

Step 2: Find the critical value

Here C is the center of mass of an object. As the object starting sliding down

$|{mgsinα}_{o}| > |μN|$

However, $|μN| = |{μmgcosα}_{o}|$

Hence the object starts sliding down when $|{mgsinα}_{o}| > |{μmgcosα}_{o}|$

$|{sinα}_{o}| > |{μcosα}_{o}|$

Assume that the inclined plain is not vertical. Hence, $\cos α_{o} \neq 0$ . Then

$|\frac{{sinα}_{o}}{{cosα}_{o}}| = |{tanα}_{o}|$

Both $μ$ and ${tanα}_{o}$ are positive,

so $|{tanα}_{o}| = {tanα}_{o}$ and $|μ| = μ$

Therefore, the object starts sliding down when role="math" localid="1664209459647" ${tanα}_{o} > μ$ .

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