Calculate Tp and Tq. (P is mid point)If you don't know just leave the question.. Created by tanviii05. Physics

Calculate Tp and Tq. (P is mid point)If you don't know just leave the question....

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tanviii05 @tanviii05

June 2021 1 632 Report

Calculate Tp and Tq. (P is mid point)
If you don't know just leave the question.

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snehitha2

Verified answer

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$}\put(-0.5,0){\vector(0,1){1.5}}\put(-1.1,1){\sf m'a}\end{picture}$