8.13 String and spring (Page 23/3)

Physics for k-12 Page 1 / 3

String and spring both transmit force from one point to another in a body system. In addition, the string facilitates change of direction of the force, whereas spring facilitates storage of force. Our treatment of these elements is restricted to ideal elements having no mass in comparison to other parts of the body sytem.

String and spring are very close in their roles and functionality in a mechanical arrangement. Main features of two elements in their ideal forms are :

String : Inextensible, mass-less : Transmits force
Spring : Extensible and compressible, mass – less : Transmits and stores force

We have already discussed the basics of these two elements in the module titled " Analysis framework for laws of motion ". Here, we shall examine their behaviour as part of a body system and shall also discuss similarities and differences between two elements.

String

The elements of a body system connected by a string have common acceleration. Consider the body system comprising of two blocks. As the system comprises of inextensible elements, the body system and each element, constituting it, have same magnitude of acceleration (not acceleration as there is change of direction), given by :

Connected system involving string

The elements of a body system connected by a string have common acceleration.

$\begin{array}{l} a = \frac{F}{m_{1} + m_{2}} \end{array}$

In the process, string transmits force undiminished for the ideal condition of being “inextensible” and “mass-less”. This is the basic character of string. We must however be careful in stating what we have stated here. If the ideal condition changes, then behavior of string will change. If string has certain “mass”, then force will not be communicated “undiminished”. If string is extensible, then it behaves like an extensible spring and accelerations of the string will not be same everywhere.

Inextensible and mass-less string

Determination of common acceleration and tension in the string is based on application of force analysis in component forms. In general, force analysis for each of the block gives us an independent algebraic relation. It means that we shall be able to find as many unknowns with the help of as many equations as there are blocks in the body system.

Individual string has single magnitude of tension through out its length. On the other hand, if the systems have different pieces of strings (when there are more than two blocks), then each piece will have different tensions.

Problem : Two blocks of masses “ $m_{1}$ ” and “ $m_{2}$ ” are connected by a string that passes over a mass-less pulley as shown in the figure. Neglecting friction between surfaces, find acceleration of the block and tension in the string.

Connected system involving string

The elements of a body system connected by a string have common acceleration.

Solution : Since there is no friction, the block of mass " $m_{1}$ " will be pulled to right by the tension in the string (T). On the other hand, block of mass " $m_{2}$ " will move down under net force in vertical direction.

Two blocks are connected by a taut string. Hence, magnitude of accelerations of the bodies are same. Let us assume that the system moves with magnitude of acceleration “a” in the directions as shown. It should be understood here that a string has same magnitude of acceleration - not the acceleration as string in conjunction with pulley changes the direction of motion and hence that of acceleration.

The external forces on " $m_{1}$ " are (i) weight of block, $m_{1} g$ , (ii) tension, T, in the string and (iii) Normal force on the block, N. We need not consider analysis of forces on $m_{1} g$ in y-direction as forces are balanced in that direction. In x-direction,

$\begin{array}{l} \sum F_{x} = T = m_{1} a \end{array}$

On the other hand, the external forces on " $m_{2}$ " are (i) weight of block, $m_{2} g$ and (ii) tension, T, in the string.

$\begin{array}{l} \sum F_{y} = m_{2} g - T = m_{2} a \end{array}$

Combining two equations, we have :

$\begin{array}{l} a = \frac{m_{2} g}{m_{1} + m_{2}} \end{array}$

and

$\begin{array}{l} \Rightarrow T = m_{1} a = \frac{m_{1} m_{2} g}{m_{1} + m_{2}} \end{array}$

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Source: OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175

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