Question

Calculate Tp and Tq. (P is mid point)
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Answer 1

Question:

Calculate $\sf T_P \ and \ T_Q$. (P is the mid point)

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines \put(0,0){\line(1,0){3}}\put(1.5,0){\line(0,-1){0.5}}\put(1,-0.5){\line(1,0){1}}\put(2,-0.5){\line(0,-1){0.5}}\put(2,-1){\line(-1,0){1}}\put(1,-1){\line(0,1){0.5}}\put(1,-0.9){\small 1.9 kg}\put(1,-0.35){\bf Q}\put(1.5,-0.25){\circle*{0.15}}\put(1.5,-1){\line(0,-1){1.5}}\put(1,-1.9){\bf P}\put(1.3,-1.8){\line(1,0){0.4}}\put(1.8,-1.9){\sf 0.2 kg}\put(1,-2.5){\line(1,0){1}}\put(2,-2.5){\line(0,-1){0.5}}\put(2,-3){\line(-1,0){1}}\put(1,-3){\line(0,1){0.5}}\put(1,-2.9){\small 2.9 kg}\put(3.2,-0.4){\vector(0,1){0.5}}\put(3.3,-0.3){ \sf a_0=0.2 m/s^2 }\put(3.3,-0.8){ \sf g=9.8 m/s^2 }\end{picture}$

Answer:

$\sf T_P$ = 30 N

$\sf T_Q$ = 50 N

Explanation:

The string between the block of mass 1.9 kg and the top is massless string.

Massive string is present between the two blocks.

Mass of the string, m = 0.2 kg

The system is accelerating upwards with acceleration, a = 0.2 m/s²

Tension at Q :

Q is located at a point on the massless string.

The blocks of masses 1.9 kg and 2.9 kg and the string having mass 0.2 kg are loaded by the massless string.

Hence, total mass, M = 1.9 kg + 2.9 kg + 0.2 kg = 5 kg

         $\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines \put(0,0){\circle*{0.1}}\put(0,0){\vector(0,1){2}}\put(0,0){\vector(0,-1){2}}\put(0,2.2){ \sf T_Q }\put(0,-2.2){ \sf Mg }\put(-0.5,0){\vector(0,1){1.5}}\put(-1.1,1){\sf Ma}\end{picture}$

$\sf T_Q-Mg=Ma \\\\ \sf T_Q=Ma+Mg \\\\ \sf T_Q=M(a+g)$

$\sf T_Q=5(0.2+9.8) \\\\ \sf T_Q=5(10) \\\\ \sf T_Q=50 \ N$

Therefore, tension at Q is 50 N

Tension at P :

P is located at mid point of the string between the two blocks.

Hence, the point P on the string loads the block of mass 2.9 kg and half of the mass of the string.

Mass of loaded string, m' = 2.9 + 0.2/2 = 2.9 + 0.1 = 3 kg

   $\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines \put(0,0){\circle*{0.1}}\put(0,0){\vector(0,1){2}}\put(0,0){\vector(0,-1){2}}\put(0,2.2){ \sf T_P }\put(0,-2.2){ \sf m'g }\put(-0.5,0){\vector(0,1){1.5}}\put(-1.1,1){\sf m'a}\end{picture}$

$\sf T_P-m'g=m'a \\\\ \sf T_P=m'a+m'g \\\\ \sf T_P=m'(a+g) \\\\ \sf T_P=3(0.2+9.8) \\\\ \sf T_P=3(10) \\\\ \sf T_P=30 \ N$

Therefore, tension at P is 30 N