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Laws of Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsLaws of Motion5 Marks
Question
Three blocks are connected as shown on a horizontal frictionless table, and pulled to the right with a force of $T_3 = 60N$.
If $m_1 = 10kg, m_2 = 20kg$ and $m_3 = 30kg$. Prove that $\frac{\text{T}_1}{\text{T}_2}=\frac{1}{3}.$
✓

Answer

From the forces acting along the horizontal (shown below) $T_1=m_1 a \ldots$...(i) $T_2-T_1=m_2 a \ldots$
(ii) $T_3-T_2=m_3 a \ldots$ Adding equation (i), (ii) and (iii), we get $T_3=\left(m_1+m_2+m_3\right)$ a or $a=\frac{T_3}{m_1+m_2+m_3}$

Putting various values, we get $\text{a}=\frac{60}{10+20+30}=\frac{60}{60}=1\text{ ms}^{-2}$
$\therefore T_1 = m_1a = 10N$ and $T_2 - T_1 = m_2a = (20 \times 1)N$ or $T_2 = (10 + 20)N = 30N$ Thus the required ratio $\frac{\text{T}_1}{\text{T}_2}=\frac{10}{30}=\frac{1}{3}$

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