Question

Three identical rigid circular cylinders A, B and C are arranged on smooth inclined surfaces as shown in figure. The least value of $\theta$ that prevent the arrangement from collapse is?

• A. $ta{n}^{-1}\left(\frac{1}{4\sqrt{3}}\right)$
• B. $ta{n}^{-1}\left(\frac{1}{2\sqrt{3}}\right)$
• C. $ta{n}^{-1}\left(\frac{1}{3\sqrt{3}}\right)$
• D. $ta{n}^{-1}\left(\frac{1}{4\sqrt{3}}\right)$

Answer 1

The correct option is C

$ta{n}^{-1}\left(\frac{1}{3\sqrt{3}}\right)$

Draw the Free Body Diagram of C, A and B. well the Free Body Diagram of A & B will be same as they are placed symmetrically.

Free Body Diagram of C

Well if you are thinking why the angle ${30}^{\circ }$ between Normal and weight. Then lets go back to ${10}^{th}$ class geometry.

3 symmetric circles touching each other - if you join the centres of these 3 circles they form an equilateral triangle and the sides exactly pass through the points where the circles touch.

If you draw XO as the tangent at the point of contact of the circle A & B then line AB is the normal to it as we know that a tangent drawn from the point where the radius meets the circumference is perpendicular to the given radius.

So this line is the one on which our normal will act as our normal has to be perpendicular to the surface at the point of contact

Now, go back to the question.

C is at equilibrium

So no net external force

$\sum f_x=\sum f_y=0$

$(N_{A}sin{30}^{\circ }=N_{B}sin{30}^{\circ }=mg$ ----------------- (I)

$N_A=N_B$

Substituting in equation (I)

$2N_{A}cos{30}^{\circ }=mg$

$N_A=\frac{mg}{\sqrt{3}}$ ---------------------(III)

Free Body Diagram of A

A is also in equilibrium

So $\sum f_x=\sum f_y=0$

$\Rightarrow mg+N_{c}cos{30}^{\circ }=Ncos\theta$ ----------- (IV)

Now if I push you with a force of 10N then according to 3rd law you will push me back with 10 N in opposite direction.

So here if A is giving a normal of $N_A$ is giving a normal of $N_A=\frac{mg}{\sqrt{3}}$ to C then C will give back the same normal $N_c$ on A just in opposite direction

$\Rightarrow N_A=N_C=\frac{mg}{\sqrt{3}}$

$\Rightarrow$ substituting in equation (IV) w get

$mg+\frac{mg}{\sqrt{3}}\times \frac{\sqrt{3}}{2}=Ncos\theta$

$\frac{3mg}{2}=Ncos\theta$ ------------------ (V)

$\sum f_x=0$

$N_{c}sin{30}^{\circ }=Nsin\theta$

$\frac{mg}{\sqrt{3}}\times \frac{1}{2}=Nsin\theta$ ------------------- (VI)

Divide (VI) with (V) we get

$tan\theta =\frac{\frac{mg}{2\sqrt{3}}}{\frac{3mg}{2}}=\left(\frac{1}{3\sqrt{3}}\right)$

$\theta =ta{n}^{-1}\left(\frac{1}{3\sqrt{3}}\right)$