Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

The tension in a string holding a solid block below the surface of a liquid (of density greater than the block) is $T_{0}$ when the container (Fig. 14-57) is at rest. When the container is given an upward acceleration of $0 .250 g$ , what multiple of $T_{0}$ gives the tension in the string?

The tension in the string is $1.25 T_{5}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Listing the given quantities

The tension in the string when the container is at rest $(T_{0})$ .

Understanding the concept of pressure

Newton’s second law states that the net force $\vec{F}$ on a body with mass m is related to the body’s acceleration $\vec{a}$ by,

$\vec{F} = m \vec{a}$

Using the condition for equilibrium $F_{net} = 0$ ,we can write an equation for tension $T_{0}$ in the string in terms of weight and buoyancy force $F_{b}$ when the container is at rest. Then we can write the force equation using Newton’s second law for a block moving up with acceleration a. From these two equations, we can write T in terms of $T_{0}$

Formulae:

At equilibrium, $F_{net} = 0$

Force of buoyancy $F_{b} = ρ_{l} V g$

Calculating the tension in the string

Free body diagram when the container is not accelerating:

When the container is at rest, the block is at equilibrium.

At equilibrium,

$F_{net} = 0$

From Free body diagram,

$\begin{aligned} F_{b} - m g - T_{0} = 0 \\ ρ V g - m g - T_{0} = 0 \\ T_{0} = ρ V g - m g \end{aligned}$

Free body diagram when container is accelerating:

Applying Newton’s law, we get

$\begin{array}{r} F_{b} - m g - T = m a \\ ρ V (g + a) - m g - T = m a \\ ρ V g + ρ V a - m g - T = m a \end{array}$

Putting $ρ V g - m g = T_{0}$ from equation (i) in it, we get

$\begin{aligned} T_{0} + ρ V a - T = m a \\ T = T_{0} + ρ V a - m a \\ = T_{0} + (ρ V - m) a \end{aligned}$

So,

$\begin{aligned} \begin{aligned} T & = T_{0} + \frac{a}{g} (ρ V g - m g) \\ = T_{0} + \frac{a}{g} T_{0} \\ = T_{0} (1 + \frac{a}{g}) \end{aligned} \\ For a = 0.250 g \\ T - T_{0} (1 + \frac{0.250 g}{g}) \\ = 1.25 T_{0} \end{aligned}$

The tension in the string is $1.25 T_{0}$ .

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Fundamentals Of Physics

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Short Answer

The tension in a string holding a solid block below the surface of a liquid (of density greater than the block) is $T_{0}$ when the container (Fig. 14-57) is at rest. When the container is given an upward acceleration of $0 .250 g$ , what multiple of $T_{0}$ gives the tension in the string?

Step by Step Solution

Listing the given quantities

Understanding the concept of pressure

Calculating the tension in the string

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Fundamentals Of Physics

Answers without the blur.

Short Answer

The tension in a string holding a solid block below the surface of a liquid (of density greater than the block) is T0 when the container (Fig. 14-57) is at rest. When the container is given an upward acceleration of 0.250g, what multiple of T0 gives the tension in the string?

Step by Step Solution

Listing the given quantities

Understanding the concept of pressure

Calculating the tension in the string

Most popular questions for Physics Textbooks

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The tension in a string holding a solid block below the surface of a liquid (of density greater than the block) is $T_{0}$ when the container (Fig. 14-57) is at rest. When the container is given an upward acceleration of $0 .250 g$ , what multiple of $T_{0}$ gives the tension in the string?