Question

Three identical billiard balls each of mass 1 kg are placed
as shown. The upper ball is smooth. The minimum
coefficient of friction between lower balls and ground, so
that the balls remain stationary is
8​

Answer 1

Answer:

Answer is 1/3√3

Explanation:

Best of luck for your AIATS

Answer 2

Concept Introduction:-

It might resemble a word or a number representation of the quantity's arithmetic value.

Given Information:-

We have been given that three identical billiard balls each of mass $1$ kg

To Find:-

We have to find that The minimum coefficient of friction between lower balls and ground, so that the balls remain stationary is

Solution:-

According to the problem

$\theta=30^{\circ}$

Using the symmetry $f_{1}=F_{2}$ the $y$ component of total farce of upper hall must be equal to the weight of the ball, Therefore,

$f_{1} \cos \theta+f_{2} \cos \theta=m g$

$\begin{aligned}&2 f\left(a_{1} 30=(2 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^{2}\right)\right. \\&F \times 2 \times \frac{\sqrt{3}}{2}=10 \\&F=\frac{10}{\sqrt{3}} \mathrm{~N}\end{aligned}$

The fare in $y$ direction on ball $2$ by ball $3$ is $f_{y}=f \cos \theta=\frac{10}{\sqrt{3}} \cos 30=\frac{10}{\sqrt{3}} \times \frac{\sqrt{3}}{2}=5 \mathrm{~N}$

$F_{y}=5 \mathrm{~N}$

Net force in $y$ direction including the weight of the ball $2$. $f_{y}=m g+5=1 \times 10+5=15 \mathrm{~N}\\f_{y}=\mathrm{~N}=15 (2)$

Force on ball $2$ in $x$ direction by all $3$,

$f_{x}=f \sin 30=\frac{10}{\sqrt{3}} \times \frac{1}{2}=\frac{5}{\sqrt{3}}-\text { (2) }$

To balance the system and maintain must be equal to the fare in $x$ re position friction force

$f_{s}=f_{x}$

$\mu F_{y}=F_{x}$

Here $\mu$ is the friction coefficient.

- using equation. $\mu \cdot 15=\frac{5}{\sqrt{3}}$

$(2)$ and $(3)$,

$\begin{aligned}&\mu=\frac{5}{15 \sqrt{3}}=\frac{1}{3 \sqrt{3}} \\&\mu=\frac{1}{3 \sqrt{3}}\end{aligned}$

Final Answer:-

The minimum coefficient of friction between lower balls and ground, so

that the balls remain stationary is $\frac{1}{3 \sqrt{3}}$.

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