A trolley 'A' of mass 2kg is connected to another trolley 'B' of mass

1.	A trolley 'A' of mass 2kg is connected to another trolley 'B' of mass 10 kg withthe help of mass less inextensible string passing over a Light friction less pulley fixed as shown in figure. Find the acceleration of thesystem of trolleys A & B.(Assume no friction, g = 10ms?, sin 53° = 45, sin 37° = 3/5)10
Answer» $\displaystyle\large\boxed{\sf{a=\dfrac{11}{3}\ m\ s^{-2}}}$ LET the acceleration act rightwards, MAKING the trolley 'A' go upward and the trolley 'B' go downward (Mass of 'B' is GREATER than that of 'A'). For simple SOLUTION, the normal reaction acting on the blocks are not considered. Free body diagram of A:- $\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(-3,4){2}{\line(3,4){5}}\multiput(0,0)(5.1,6.8){2}{\line(-3,4){3}}\put(-0.4,4.1){$\sf{A}$}\put(2.55,3.4){\vector(0,-1){10}}\put(-1,-10){$\sf{20\ N}$}\put(3.5,8.8){\vector(3,4){7}}\put(11,17){$\sf{T}$}\put(12,2){\vector(3,4){10}}\put(18,5){$\sf{a}$}\put(2.55,3.4){\vector(3,-4){5}}\put(2.55,3.4){\vector(-3,-4){5}}\put(3,-3.7){\tiny\text{$53^{\circ}$}}\end{picture}$ From this, $\displaystyle\longrightarrow\sf{T-20\sin53^{\circ}=2a}$ $\displaystyle\longrightarrow\sf{T-20\times\dfrac{4}{5}=2a}$ $\displaystyle\longrightarrow\sf{T-16=2a\quad\quad\dots(1)}$ Free body diagram of B:- $\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(-8,6){2}{\line(3,4){3}}\multiput(0,0)(3,4){2}{\line(-4,3){8}}\put(-4,4){$\sf{B}$}\put(-4,3){\vector(0,-1){10}}\put(-4,3){\vector(4,-3){6}}\put(-4,3){\vector(-3,-4){5.6}}\put(-8,-11){$\sf{100\ N}$}\put(-8,-4){\tiny\text{$37^{\circ}$}}\put(-6.6,7.8){\vector(-4,3){8}}\put(-17.8,13){$\sf{T}$}\put(-5,16){\vector(4,-3){10}}\put(0,14){$\sf{a}$}\end{picture}$ From this, $\displaystyle\longrightarrow\sf{100\sin37^{\circ}-T=10a}$ $\displaystyle\longrightarrow\sf{100\times\dfrac{3}{5}-T=10a}$ $\displaystyle\longrightarrow\sf{60-T=10a\quad\quad\dots(2)}$ Adding (1) and (2), $\displaystyle\longrightarrow\sf{60-T+T-16=2a+10a}$ $\displaystyle\longrightarrow\sf{44=12a}$ $\displaystyle\longrightarrow\sf{\underline{\underline{a=\dfrac{11}{3}\ m\ s^{-2}}}}$

A trolley 'A' of mass 2kg is connected to another trolley 'B' of mass 10 kg withthe help of mass less inextensible string passing over a Light friction less pulley fixed as shown in figure. Find the acceleration of thesystem of trolleys A & B.(Assume no friction, g = 10ms?, sin 53° = 45, sin 37° = 3/5)10

Answer»

$\displaystyle\large\boxed{\sf{a=\dfrac{11}{3}\ m\ s^{-2}}}$

LET the acceleration act rightwards, MAKING the trolley 'A' go upward and the trolley 'B' go downward (Mass of 'B' is GREATER than that of 'A').

For simple SOLUTION, the normal reaction acting on the blocks are not considered.

Free body diagram of A:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(-3,4){2}{\line(3,4){5}}\multiput(0,0)(5.1,6.8){2}{\line(-3,4){3}}\put(-0.4,4.1){$\sf{A}$}\put(2.55,3.4){\vector(0,-1){10}}\put(-1,-10){$\sf{20\ N}$}\put(3.5,8.8){\vector(3,4){7}}\put(11,17){$\sf{T}$}\put(12,2){\vector(3,4){10}}\put(18,5){$\sf{a}$}\put(2.55,3.4){\vector(3,-4){5}}\put(2.55,3.4){\vector(-3,-4){5}}\put(3,-3.7){\tiny\text{$53^{\circ}$}}\end{picture}$

From this,

$\displaystyle\longrightarrow\sf{T-20\sin53^{\circ}=2a}$

$\displaystyle\longrightarrow\sf{T-20\times\dfrac{4}{5}=2a}$

$\displaystyle\longrightarrow\sf{T-16=2a\quad\quad\dots(1)}$

Free body diagram of B:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\multiput(0,0)(-8,6){2}{\line(3,4){3}}\multiput(0,0)(3,4){2}{\line(-4,3){8}}\put(-4,4){$\sf{B}$}\put(-4,3){\vector(0,-1){10}}\put(-4,3){\vector(4,-3){6}}\put(-4,3){\vector(-3,-4){5.6}}\put(-8,-11){$\sf{100\ N}$}\put(-8,-4){\tiny\text{$37^{\circ}$}}\put(-6.6,7.8){\vector(-4,3){8}}\put(-17.8,13){$\sf{T}$}\put(-5,16){\vector(4,-3){10}}\put(0,14){$\sf{a}$}\end{picture}$