A uniform meter scale of mass m is suspended | vedclass

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(D) Let the length of the meter scale be $L = 1 \ m$. The weight of the scale $mg$ acts at its center of mass,i.e.,at $0.5 \ m$ from end $A$. The addi

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$7/13$

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(D) Let the length of the meter scale be $L = 1 \ m$. The weight of the scale $mg$ acts at its center of mass,i.e.,at $0.5 \ m$ from end $A$. The additional mass $m$ is placed at $0.8 \ m$ from end $A$.Taking the torque about point $A$ for rotational equilibrium:$\sum 	au_A = 0$$(mg 	imes 0.5) + (mg 	imes 0.8) = T_2 	imes 1$$0.5 \ mg + 0.8 \ mg = T_2$$T_2 = 1.3 \ mg$For translational equilibrium,the sum of vertical forces is zero:$T_1 + T_2 = mg + mg = 2 \ mg$$T_1 + 1.3 \ mg = 2 \ mg$$T_1 = 0.7 \ mg$Now,the ratio of the tensions is:$\frac{T_1}{T_2} = \frac{0.7 \ mg}{1.3 \ mg} = \frac{7}{13}$

$A$ uniform meter scale of mass $m$ is suspended by two vertical strings at its ends as shown in the figure. $A$ mass $m$ is placed at the $80 \ cm$ mark. Find the ratio of the tension forces in the strings $(T_1/T_2)$.

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