Question

Two particles A and B, each of mass m are kept stationary by applying a horizontal force F = mg on particle B as shown in figure. Then:

${\text{A}}{\text{. tan}}\beta {\text{ = 2tan}}\alpha $
${\text{B}}{\text{. 2}}{{\text{T}}_1} = 5{{\text{T}}_2}$
${\text{C}}{\text{. }}\sqrt 2 {{\text{T}}_1} = \sqrt 5 {{\text{T}}_2}$
${\text{D}}{\text{.}}$ None of these

Answer 1

Hint: At first draw a free body diagram for particles A and B and then resolve the forces into their components along X- axis and Y- axis.
Formula used - ${T_1}\sin \alpha = {T_2}\sin \beta $ , ${T_1}\cos \alpha = {T_2}\cos \beta + mg$ , ${T_2}\cos \beta = mg$ , ${T_2}\sin \beta = mg$

Complete Step-by-Step solution:
Let us first draw the free body diagram (FBD) for both the particles A and B.
Refer the image shown (FBD)-

Here $a = \alpha ,b = \beta $ .
Now resolving these forces into their components along X- axis and Y- axis, we get-
From the FBD of A,
${T_1}\sin \alpha = {T_2}\sin \beta - (1)$
${T_1}\cos \alpha = {T_2}\cos \beta + mg - (2)$
Now from the FBD of B,
${T_2}\cos \beta = mg - (3)$
${T_2}\sin \beta = mg - (4)$
now dividing equation 4 by equation 3 we get-
$\dfrac{{{T_2}\sin \beta }}{{{T_2}\cos \beta }} = \dfrac{{mg}}{{mg}}$
So, we get- $\tan \beta = 1$
Using equation (4) in (1) and equation (3) in (2),
$T\sin \alpha = mg$ and $T\cos \alpha = 2mg$
So, dividing these two we get-
$\tan \alpha = \dfrac{1}{2}$
We can also write,
 $
  2\tan \alpha = 1 \\
   \Rightarrow 2\tan \alpha = \tan \beta \{ \because \tan \beta = 1\} \\
$
It means option A is correct.
Now using $\alpha $ and $\beta $ values in (1) we get-
${T_1}\sin \alpha = {T_2}\sin \beta $
$\sqrt 2 {{\text{T}}_1} = \sqrt 5 {{\text{T}}_2}$ (option C)
Therefore, options A and C are correct.

Note – Whenever such types of questions appear then always write the things given in the question and then by drawing the FBD of A and B solve the question, as done in the solution, the forces are resolved into components and then we have found the correct options by using the equations mentioned in the solution.