Question

find
acceleration of B​

Answer 1

Given :

Two blocks of masses 3m and m are connected as shown in the figure.

To Find :

Acceleration of block B.

Solution :

❒ This question is based on the concept of constraint motion.

• First of all we need to find relation between acceleration of both blocks.

Let acceleration of block A (3m) be a and that of block B (m) be a'.

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\framebox(2.5,1){\bf 3m}}\put(4,0){\framebox(1,1){\bf m}}\put(1.3,0){\vector(0,-1){1.5}}\put(0.6,1){\vector(0,1){1}}\put(1.8,1){\vector(0,1){1}}\put(5,0.5){\vector(1,0){1}}\put(1,-2){\bf 3mg}\put(0.4,2.3){\bf 2T}\put(1.65,2.3){\bf T}\put(6.3,0.4){\bf T}\put(-1,2){\vector(0,-1){2.5}}\put(3.7,1.7){\vector(1,0){2.5}}\put(-1.15,-1){\bf\large a}\put(6.5,1.6){\bf\large a'}\end{picture}$

» Tension force on block A = 3T

» Tension force on block B = T

As per constraint motion equation;

$\sf:\implies\:\sum (T\cdot a)=0$

$\sf:\implies\:T_1\cdot a_1+T_2\cdot a_2=0$

$\sf:\implies\:(-3T)(a)+(T)(a')=0$

$\sf:\implies\:3T\times a=T\times a'$

$\bf:\implies\:a=\dfrac{a'}{3}\:\dots\:(I)$

★ Force equation of block A :-

$\sf:\implies\:(3m)g-3T=(3m)a$

$\sf:\implies\:mg-T=ma$

$\bf:\implies\:T=mg-ma\:\dots\:(II)$

★ Force equation of block B :-

$\sf:\implies\:T=ma'$

From the (II) equation,

$\sf:\implies\:mg-ma=ma'$

$\sf:\implies\:g-a=a'$

From the (I) equation,

$\sf:\implies\:g-\dfrac{a'}{3}=a'$

$\sf:\implies\:a'+\dfrac{a'}{3}=g$

$\sf:\implies\:\dfrac{4a'}{3}=g$

$:\implies\:\underline{\boxed{\bf{\purple{a'=\dfrac{3g}{4}\:ms^{-2}}}}}$