A container partially filled with water is moved horizontally with acceleration $$a=g3$$. A small wooden ball of mass m is tied to the bottom of the container using a string. The ball remains inside water with the string inclined at an angle θ to the horizontal. Assuming that the density of ball is half the density of water, if, then $$m=310kg$$ find the tension in the string (in$$ $$N). (Take$$ g $$=$$ $$10$$ $$m/s2)

Question

A container partially filled with water is moved horizontally with acceleration $$a=g3$$. A small wooden ball of mass m is tied to the bottom of the container using a string. The ball remains inside water with the string inclined at an angle θ to the horizontal. Assuming that the density of ball is half the density of water, if, then $$m=310kg$$ find the tension in the string (in$$ $$N). (Take$$ g $$=$$ $$10$$ $$m/s2)

Infinity Learn · Accepted Answer

As the container is accelerating the liquid surface will be inclined with horizontal such that

$\tan θ = \frac{a}{g} = \frac{g / 3}{g} = \frac{1}{3}$

Hence $\sin θ = \frac{1}{\sqrt{10}} and \cos θ = \frac{3}{\sqrt{10}}$

In this case the bouncy force on the ball will be natural to water surface. The magnitude of bouncy force

$B = V_{dis} ρ_{w} g_{eff}$

$\Rightarrow B = (\frac{m}{ρ}) ρ_{w} \sqrt{g^{2} + a^{2}} = 2 m \sqrt{g^{2} + \frac{g^{2}}{9}}$

which gives $B = \frac{2 \sqrt{10}}{3} mg$

The force acting on the ball are shown in F.B.D. (w.r.t. container)

As the ball is in equilibrium with respect to container, then

$B = T + mgcos θ + masin θ$

$\frac{2 \sqrt{10}}{3} mg = T + mg \cdot \frac{3}{\sqrt{10}} + m (\frac{g}{3}) \cdot \frac{1}{\sqrt{10}}$

$\Rightarrow T = \frac{\sqrt{10}}{3} mg = \frac{\sqrt{10}}{3} \times 3 \sqrt{10} \times 10 = 100 N$

Pressure in a fluid-mechanical properties of fluids

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