Suggested languages for you:

Americas

Europe

Q86P

Expert-verified
Found in: Page 412

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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

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

See the step by step solution

## Listing the given quantities

The tension in the string when the container is at rest $\left({T}_{0}\right)$.

## Understanding the concept of pressure

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

${\mathbf{}}\stackrel{\mathbf{\to }}{\mathbf{F}}{\mathbf{=}}{\mathbit{m}}\stackrel{\mathbf{\to }}{\mathbf{a}}$

Using the condition for equilibrium ${{\mathbit{F}}}_{{\mathbf{\text{net}}}}{\mathbf{=}}{\mathbf{0}}$,we can write an equation for tension ${{\mathbit{T}}}_{{\mathbf{0}}}$ in the string in terms of weight and buoyancy force ${{\mathbit{F}}}_{{\mathbf{b}}}$ 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 ${{\mathbit{T}}}_{{\mathbf{0}}}$

Formulae:

At equilibrium, ${F}_{\text{net}}=0$

Force of buoyancy ${F}_{b}={\rho }_{l}Vg$

## 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,

${F}_{\text{net}}=0$

From Free body diagram,

$\begin{array}{r}{F}_{b}-mg-{T}_{0}=0\\ \rho Vg-mg-{T}_{0}=0\\ {T}_{0}=\rho Vg-mg\end{array}$

Free body diagram when container is accelerating:

Applying Newton’s law, we get

$\begin{array}{r}{F}_{b}-mg-T=ma\\ \rho V\left(g+a\right)-mg-T=ma\\ \rho Vg+\rho Va-mg-T=ma\end{array}$

Putting $\rho Vg-mg={T}_{0}$ from equation (i) in it, we get

$\begin{array}{r}{T}_{0}+\rho Va-T=ma\\ T={T}_{0}+\rho Va-ma\\ ={T}_{0}+\left(\rho V-m\right)a\end{array}$

So,

$\begin{array}{r}\begin{array}{rl}T& ={T}_{0}+\frac{a}{g}\left(\rho Vg-mg\right)\\ & ={T}_{0}+\frac{a}{g}{T}_{0}\\ & ={T}_{0}\left(1+\frac{a}{g}\right)\end{array}\\ \text{For}a=0.250g\\ T-{T}_{0}\left(1+\frac{0.250g}{g}\right)\\ \phantom{\rule{1em}{0ex}}=1.25{T}_{0}\end{array}$

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