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Q86P

Expert-verifiedFound in: Page 412

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 ${{\mathit{T}}}_{{\mathbf{0}}}$ 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 ${{\mathit{T}}}_{{\mathbf{0}}}$ gives the tension in the string?**

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

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

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

**${\mathbf{}}\overrightarrow{\mathbf{F}}{\mathbf{=}}{\mathit{m}}\overrightarrow{\mathbf{a}}$ **

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

Formulae:

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

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

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

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

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

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

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

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

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