Many common situations involve tying two objects by using a rope or a string. When the connecting string is taut, it acts as a stretched spring and pulls in the the two bodies at the two ends. We say that the taut string provides a tension force on the two bodies. We will denote the magnitude of the tension force by \(F_T\) and sometimes just by \(T\text{.}\)
Tension develops as a reaction to some other external force such as gravity or pull on the two bodies connected by the string. Consider hanging an object A from ceiling B as shown in Figure 6.73. The gravity pulls on A, which causes the string to become taut with some tension of magnitude \(F_T\text{.}\) In the figure, \(F_T\) acts upward on A, opposing the external force that caused it, while this same force acts downward when acts on B, which is taken to be “fixed”.
Figure6.73.
Unlike spring force, we do not have any formula for the magnitude of the tension force. If you have memorized formulas such as \(F_T = mg\text{,}\) then you need to forget that since there is no such law although in Figure 6.73if A is not accelerating, then balancing the forces on A does give \(F_T = m_A g\text{.}\) In general, magnitude of tension is obtained by setting up Newton’s equation of motion. When we study elasticity, we will be able to give a formula for tension in terms of Young’s modulus using Hooke’s law. But, for now, we will just represent tension by a unknown symbol.
Example6.74.Supporting a hanging mass.
A 50-kg block is hanging from the ceiling by a light rope. (a) Find the tension in the string. Now, suppose you place your hand and push vertically up on the block from underneath it by a force of 100 N. (b) What is the tension in the rope? (c) What will be the tension in the rope if you were to pull the block down with 100 N? (d) What will happen if you push up by a force of 600-N?
Since \(a=0\text{,}\) forces are balanced. Furthermore, forces here are all along the same line. Using any axis along the line of forces, will immediately give
\begin{equation*}
T - mg = m \times 0 = 0.
\end{equation*}
Again, since \(a=0\text{,}\) forces are balanced. Furthermore, forces here are all along the same line. Using any axis along the line of forces, will immediately give
\begin{equation*}
F + T' - mg = m \times 0 = 0.
\end{equation*}
This is just like part (b) except that now the applied force is pointed down. Again, since \(a=0\text{,}\) forces are balanced. Furthermore, forces here are all along the same line. Using any axis along the line of forces, will immediately give
Now, we cannot assume \(a=0\text{.}\) If the string remained taut, tension on the block must point up. Therefore, the value of \(T\) must be positive.
Let \(y\)-axis be pointed up and we look at the \(y\)-component of net force.
\begin{equation*}
F + T - mg = 600 + T -500 = T + 100.
\end{equation*}
If \(a_y\) were zero, setting this to zero would give \(T = -100\text{ N}\text{,}\) which is not possible if the string is taut. Hence, \(a_y\ne 0\) and string is loose with tension \(T=0\text{.}\) Actually, setting this to \(ma_y\) gives
Bodies connected by strings are common occurence. A great example of a conected system is Atwood machine shown in Figure 6.79, where two blocks are tied together by a string that goes over a pulley. The string connects the dynamics of three bodies in this system: the two blocks and the pulley. By requiring that the length of the string do not change during the motion, we immediately get that magnitude of displacement of the two blocks must be equal. Furthermore, if the string is not to slide over the rotating pulley, then the angle by which pulley should rotate will also be related to the magnitude of displacement.
If string does not slide on the pulley surface due to sufficient static friction, the strings on the two sides of the pulley may have two different magnitudes \(T_1 \) and \(T_2\) with \(T_1 \ne T_2\text{.}\) However, if pulley’s mass is negligible compared to the masses of the blocks, we can assume \(T_1 = T_2\) and represent them with the same symbol \(T\text{.}\) We will study the dynamics of coupled systems in another section.
Figure6.79.
Example6.80.
In Figure 6.81, there are tension forces between the ring and the block \(m\) and the ring and the support \(S\text{.}\) The tensions in the two sides of the ring may be different: \(T_1 \ne T_2\text{.}\) Furthermore, while there is one tension force acting on mass \(m\) and another one acting on the support \(S\text{,}\) both tension forces act on the ring \(R\text{.}\) The net tension force on the ring is the vector sum of the two tension forces acting on it.
Figure6.81.The tension in the string acts on three objects, the mass \(m\text{,}\) the ring \(R\) and the support \(S\text{.}\) The tension force between mass \(m\) and ring \(R\) may be different from the tension force between the ring and the support \(R\) although the same wire connects all three objects. In (b) the tension force on mass \(m\) is shown, in (c) the tension force on the support \(S\) is shown, and in (d) the tension forces on ring \(R\) is shown. The net tension force on ring \(R\) is the vector sum of the two tensions.
Exercises6.16.2Exercises
1.Tension Changing Direction Via a Pulley.
(a) In Figure 6.82, how much will the spring stretch if \(T_1 = T_2\) and there is no friction between block 1 and the surface of the table? If \(m_1=10\,\text{kg}\) and (b) \(m_2=20\text{ kg}\text{,}\) what is the value of tension?
Figure6.82.
Hint.
Look at forces on each mass separately.
Answer.
(a) \(m_2 g/k\text{.}\)
2.Varying Tension in a Hanging Chain.
A long chain consists of links of mass \(m\) each. The chain is hung from a tower. What is the tension in the chain at the \(n^{\text{th}}\) link from the bottom? Answer.
\(n m g\text{.}\)
Solution.
There are two forces on the \(n^{\text{th}}\) link, the weight of \(n\) links below, which is pointed down and the tension \(T_n\) at the site pointed up.
\begin{equation*}
T_n = n m g\ \ \ ( n = 1, 2,\cdots, N ).
\end{equation*}
