7.0 Tension Force

Rules of applying tension force on a body or string

Whenever we apply tension force on a body, we follow certain rules:

1. The tension force should be along the length of the body.

2. The tension force should be always applied away from the body.

Types of tension force

1. Extensional tension

2. Compressional tension

Note:

• Tension in a string or in a chain is always extensional
• Tension in a body or any rigid body can be both extensional and compressional both.

Properties of an ideal string

1. The mass of an ideal spring is zero

2. The net acting force on a string is always zero, $$\begin{equation} \begin{aligned} {F_{net}} = ma \\ {F_{net}} = 0.\;a\quad \left( {{\text{massless}}} \right) \\ {F_{net}} = 0 \\\end{aligned} \end{equation} $$

3. Tension at all points in the string is same

4. Ideal string is inextensible, which means that the length of the ideal string remains constant

Question 3. Find out the tension in the both the string as shown in the figure.

Solution: Suppose the whole system moves with an acceleration $a$,

We will consider only those forces, which affect the motion. So, only horizontal forces are considered.

$$\begin{equation} \begin{aligned} {T_1} = {m_1}a\quad ...(i) \\ F - {T_1} = {m_2}a\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} F = \left( {{m_1} + {m_2}} \right).\;a \\ a = \left( {\frac{F}{{{m_1} + {m_2}}}} \right)\quad ...(iii) \\\end{aligned} \end{equation} $$

From equation $(i)$ & $(iii)$ we get, $${T_1} = \left( {\frac{{{m_1}F}}{{{m_1} + {m_2}}}} \right)$$

As the string is ideal, so the tension will be same through out. Therefore, the tension in the second string is, $${T_2} = F$$

Question 4. Two unequal forces $F_1$ and $F_2$ $\left( {{F_1} > {F_2}} \right)$ act on the blocks of masses $m_1$ and $m_2$ respectively as shown in the figure. Find the tension in the string and the acceleration of the system.

Solution: Assume that the whole system moves with an acceleration $a$.

Vertical forces do not affect the motion, so only the forces in the horizontal direction are considered.

From FBD we can write, $$\begin{equation} \begin{aligned} T - {F_2} = {m_2}a\quad ...(i) \\ {F_1} - T = {m_1}a\quad ...(ii) \\\end{aligned} \end{equation} $$

Adding equation $(i)$ and $(ii)$ we get, $$a = \left( {\frac{{{F_1} - {F_2}}}{{{m_1} + {m_2}}}} \right)\quad ...(iii)$$

From equation $(i)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} T = {F_2} + {m_1}\left( {\frac{{{F_1} - {F_2}}}{{{m_1} + {m_2}}}} \right) \\ T = \frac{{{m_1}{F_2} + {m_2}{F_1}}}{{{m_1} + {m_2}}} \\\end{aligned} \end{equation} $$