Question 261: Laws of Motion

Question

Three blocks of masses $$m_1, m_2$$ and $$m_3$$ are connected by massless strings, as shown, on a frictionless table. They are pulled with a force, $$T_3=40 \mathrm{~N}$$. If $$m_1=10 \mathrm{~kg}, m_2=8 \mathrm{~kg}$$ and $$m_3=2 \mathrm{~kg}$$, the tension $$T_2$$ will be

Accepted Answer

For equilibrium of all 3 masses,
$$a=\frac{T_3}{m_1+m_2+m_3}$$
For equilibrium of $$m_1$$ and $$m_2$$
$$T_2=\left(m_1+m_2ight) \cdot a=\frac{\left(m_1+m_2ight) T_3}{m_1+m_2+m_3}$$
$$\begin{aligned}
	ext { Given, } m_1 & =10 \mathrm{~kg}, m_2=8 \mathrm{~kg}, m_3=2 \mathrm{~kg} \
T_3 & =40 \mathrm{~N} \
	herefore \quad T_2 & =\frac{(10+8) 40}{10+8+2}=36 \mathrm{~N}
\end{aligned}$$

Answer

20 N

Answer

40 N

Answer

24 N

Answer

36 N

Laws of Motion

Three blocks of masses $m_1, m_2 and m_3 are connected by massless strings, as shown, on a frictionless table. They are pulled with a force, T_3=40 \mathrm{~N}. If m_1=10 \mathrm{~kg}, m_2=8 \mathrm{~kg} and m_3=2 \mathrm{~kg}, the tension T_2$ will be

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