Question

find the acceleration of the 6 kg mass, tension t1 and tension t2 in the strings.
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Answer 1

Let the net acceleration of the system be $\sf{a.}$ Let $\sf{g=10\ m\,s^{-2}.}$

Free Body Diagram of the 2 kg block at left is given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(8,8){\sf{2\ kg}}}\put(4,0){\vector(0,-1){10}}\put(0,-14){\sf{20\ N}}\put(4,8){\vector(0,1){10}}\put(3,19.5){ \sf{R_1} }\put(8,4){\vector(1,0){10}}\put(19,3){ \sf{T_1} }\put(0,0){\vector(-1,0){5}}\put(-9,-1){ \sf{f_1} }\put(10,15){\vector(1,0){10}}\put(14,12){\sf{a}}\end{picture}$

Since net vertical force is zero,

$\longrightarrow\sf{R_1=20\ N}$

The net horizontal force is,

$\longrightarrow\sf{T_1-f_1=2a}$

$\longrightarrow\sf{T_1-0.2\,R_1=2a}$

$\longrightarrow\sf{T_1-0.2\times20=2a}$

$\longrightarrow\sf{T_1-4=2a}$

$\longrightarrow\sf{T_1=2a+4\quad\qaud\dots(1)}$

Free Body Diagram of the other 2 kg block is given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(8,8){\sf{2\ kg}}}\put(4,0){\vector(0,-1){10}}\put(0,-14){\sf{20\ N}}\put(4,8){\vector(0,1){10}}\put(3,19.5){ \sf{R_2} }\put(8,4){\vector(1,0){10}}\put(19,3){ \sf{T_2} }\put(0,4){\vector(-1,0){10}}\put(-15,3){ \sf{T_1} }\put(0,0){\vector(-1,0){5}}\put(-9,-1){ \sf{f_2} }\put(10,15){\vector(1,0){10}}\put(14,12){\sf{a}}\end{picture}$

Since net vertical force is zero,

$\longrightarrow\sf{R_2=20\ N}$

The net horizontal force is,

$\longrightarrow\sf{T_2-T_1-f_2=2a}$

From (1),

$\longrightarrow\sf{T_2-2a-4-0.2\,R_2=2a}$

$\longrightarrow\sf{T_2-2a-4-0.2\times20=2a}$

$\longrightarrow\sf{T_2-2a-4-4=2a}$

$\longrightarrow\sf{T_2=4a+8\quad\quad\dots(2)}$

Free Body Diagram of the 6 kg block is given below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(8,8){\sf{6\ kg}}}\put(4,0){\vector(0,-1){10}}\put(0,-14){\sf{60\ N}}\put(4,8){\vector(0,1){10}}\put(3,19.5){ \sf{T_2} }\put(15,9){\vector(0,-1){10}}\put(17,4){\sf{a}}\end{picture}$

The net vertical force is given by,

$\longrightarrow\sf{60-T_2=6a}$

From (2),

$\longrightarrow\sf{60-4a-8=6a}$

$\longrightarrow\sf{10a=52}$

$\longrightarrow\underline{\underline{\sf{a=5.2\ m\,s^{-2}}}}$

I.e., the acceleration of the 6 kg block is $\bf{5.2\ m\,s^{-2}.}$

Then (1) becomes,

$\longrightarrow\sf{T_1=2\times5.2+4}$

$\longrightarrow\underline{\underline{\sf{T_1=14.4\ N}}}$

And (2) becomes,

$\longrightarrow\sf{T_2=4\times5.2+8}$

$\longrightarrow\underline{\underline{\sf{T_2=28.8\ N}}}$