Question

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?

Short answer

The tension in the string is $1.25T_5$.

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 $\overrightarrow{\mathbf{F}}$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$F_b={\rho }_{l}Vg$

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

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

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,

$\begin{array}{r}\begin{array}{rl}T& =T_0+\frac{a}{g}(\rho Vg-mg)\\ & =T_0+\frac{a}{g}{T}_0\\ & =T_0\left(1+\frac{a}{g}\right)\end{array}\\ \text{For}a=0.250g\\ T-T_0\left(1+\frac{0.250g}{g}\right)\\ =1.25T_0\end{array}$

The tension in the string is $1.25T_0$.