Question

A horizontal force of 600 N pulls two masses 10kg and 20kg lying on a frictionless surface connected by a light string.What is the tension in the string?Does the answer depend on the which mass end the pull is applied?

Answer 1

A horizontal force of $600\mathrm{N}$ pulls two masses $10\mathrm{kg}$ and $20\mathrm{kg}$ lying on a frictionless surface connected by a light string. What is the tension in the string? Does the answer depend on the which mass end the pull is applied?

Solution:

Because the string is thin and inextensible, then:

$$\vec {T} = - \vec {T ^ {\prime}}; \left| \vec {T} \right| = \left| \vec {T ^ {\prime}} \right|$$

The general case (instead of mass $10kg$ and $20kg \rightarrow m_1$ and $m_2, F = 600N$ )

Newton's second law for the mass $m_{1}$ :

$$\overline {{m _ {1} g}} + \overrightarrow {N _ {1}} + \vec {T} = m _ {1} \vec {a}$$ $$x: T = m _ {1} a$$

Newton's second law for the mass $m_2$ :

$$\overline {{m _ {2} g}} + \overrightarrow {N _ {2}} + \vec {T} + \vec {F} = m _ {2} \vec {a}$$ $$x: F - T ^ {\prime} = m _ {2} a$$ $$T = T ^ {\prime}$$

(1)in (2): $F - m_{1}a = m_{2}a$

$$a = \frac {F}{m _ {1} + m _ {2}}$$

(3)in(1): $T = m_{1}a = \frac{m_{1}F}{m_{1} + m_{2}}$

If $m_{1} = 10kg$ and $m_{2} = 20kg$ :

$$T = \frac {1 0 k g \cdot 6 0 0 N}{1 0 k g + 2 0 k g} = 2 0 0 N$$

If $m_{1} = 20kg$ and $m_{2} = 10kg$ :

$$T = \frac{20kg \cdot 600N}{10kg + 20kg} = 400N$$

Answer: if end the pull is applied to mass $20kg$, tension in the string is $T = 200N$; if end the pull is applied to mass $10kg$, tension in the string is $T = 400N$.