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Course by: Sunil Kumar Singh. E-mail the author

# String and spring

Module by: Sunil Kumar Singh. E-mail the author

Summary: 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 :

a = F m 1 + m 2 a = F m 1 + m 2

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.

#### Example 1

Problem : Two blocks of masses “ m 1 m 1 ” and “ m 2 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.

Solution : Since there is no friction, the block of mass " m 1 m 1 " will be pulled to right by the tension in the string (T). On the other hand, block of mass " m 2 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 m 1 " are (i) weight of block, m 1 g 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 m 1 g in y-direction as forces are balanced in that direction. In x-direction,

F x = T = m 1 a F x = T = m 1 a

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

F y = m 2 g - T = m 2 a F y = m 2 g - T = m 2 a

Combining two equations, we have :

a = m 2 g m 1 + m 2 a = m 2 g m 1 + m 2

and

T = m 1 a = m 1 m 2 g m 1 + m 2 T = m 1 a = m 1 m 2 g m 1 + m 2

#### Useful deduction

The result for acceleration as obtained above is significant for this type of arrangement, where blocks are pulled down by a string passing over a pulley. The acceleration as derive is equivalent to :

a = F m 1 + m 2 a = F m 1 + m 2

where F = m 2 g F = m 2 g . It suggests that we can treat the arrangement as composite mass of ( m 1 + m 2 m 1 + m 2 ) subjected to an external force “F” of magnitude “ m 2 g m 2 g ”. This analysis helps in situations where more than one block is pulled by a hanging block. Take the example as shown in the figure :

In this case,

a = F m 1 + m 2 + m 3 + m 4 = m 4 g m 1 + m 2 + m 3 + m 4 a = F m 1 + m 2 + m 3 + m 4 = m 4 g m 1 + m 2 + m 3 + m 4

T 1 = m 1 a ; T 2 = ( m 1 + m 2 ) a ; T 3 = ( m 1 + m 2 + m 3 ) a ; T 1 = m 1 a ; T 2 = ( m 1 + m 2 ) a ; T 3 = ( m 1 + m 2 + m 3 ) a ;

Important to note here is that the separate strings have different tensions.

### Mode of force application

Consider the two cases when a block is pulled down by a force in two different manners. In one case, string is pulled by another mass m 2 m 2 , hanging from it. In the second case, the string is pulled by a force equal to hanging weight ( m 2 g m 2 g ). Now the question is : whether two situations are equivalent or different ?

As worked out earlier, the acceleration in the first case, a 1 a 1 , is :

a 1 = m 2 g m 1 + m 2 a 1 = m 2 g m 1 + m 2

In the second case, let the acceleration of the block be “ a 2 a 2 ”. Then force analysis of the block on the table is :

m 1 a 2 = T m 1 a 2 = T

But,

T = m 2 g T = m 2 g

Now, combining two equations, we have :

a 2 = m 2 g m 1 a 2 = m 2 g m 1

Evidently, a 2 > a 1 a 2 > a 1 . This result is expected because applying force as weight and as force are two different things. In the first case, mass of the body applying force is also involved as against the later case, when no "body" of any mass is involved as far as application of force is concerned. The bottom line is : “do not hang a weight simply to apply force”.

### Variable acceleration of a single string

Further, the behavior of string is true to ideal string so long string is not intervened by other element, which is itself accelerating. Consider a pulley – string system, in which the pulley is accelerating as shown in the figure.

In such case, the accelerations of different parts of the string are not same. Note in the example shown above that one end of the string is fixed, but the other end of string is moving. As a matter of fact, the acceleration of block (hence that of the string at that end) is twice that of the pulley. The acceleration of the other end of the string, which is fixed, is zero. This aspect is explained in a separate module on movable pulley in the course.

### String with certain mass

The tension in the string having certain mass is not same everywhere. A part of the force is required to accelerate string as well. Usually,mass is distributed uniformly along the length of the string. We describe mass distribution in terms of "mass per unit length" and denotes the same as "λ" such that the mass of the string,"m" of length "L" is given as :

m = λL

If we take cross section at any point in the string, then the part of the string on each side forms an mass element. having mass proportional to the length of that part of the string. This aspect is brought out in the example given here.

#### Example 2

Problem : A body of mass “m” is hanging with a string having linear mass density “λ”. What is the tension at point “A” as shown in the figure.

Solution : The tension in the string is not same. The tension above “A” balances the weight of the block and the weight of the string below point “A”.

Free body diagram of point “A” Free body diagram of point “A”

Mass of the string below point A, " m s m s ", is :

m s = λ ( L - y ) m s = λ ( L - y )

The external forces at point "A" are (i) Tension, T (ii) weight of block, mg, and (iii) weight of string below A," m s g m s g ".

F y = T - mg - m s g = 0 T = ( m + m s ) g T = { m + λ ( L - y ) } g F y = T - mg - m s g = 0 T = ( m + m s ) g T = { m + λ ( L - y ) } g

## Spring

Spring is extensible and compressible. What it means that spring involves change in its length, when force is applied to it. It must be understood that the role of spring in mechanical arrangement is more significant with respect to “storage” of energy (force) rather than transmission of force. This aspect is dealt in modules related to energy. For the time being, we describe only those aspects of spring, which involve transmission of force.

As a matter of fact, spring is exactly like string – though as an element it generally gives impression of complexity. We shall, however, find that such is not the case except for minor adjustment in our perception.

### Constant acceleration

When a block (body) hangs from a spring, the body system is stationary. The spring is in extended condition for the given force system working on it. The forces at the two ends of the spring are equal. Net force on the spring is zero in the extended condition. This is same as string. There is no difference. As a matter of fact, ideal spring not only transmits force as truly as string, but also provides the magnitude of force in terms of force constant as :

F = k x 0 F = k x 0

where x 0 x 0 is the extension for a given force as applied at either end. When a block is suspended from a spring hanging from the ceiling, the spring is acted by equal forces at both ends. We calculate extension in terms of either of the two forces and not for the net force, which is zero in any case.

#### Example 3

Problem : Consider the arrangement consisting of spring in the figure. If extension in the spring is 0.01 m for the given blocks, find the spring constant.

Solution : We treat spring exactly like string. The tension in the connecting string is equal to spring force,

T = k x 0 T = k x 0

The free body diagrams of 3 kg and 2 kg blocks with assumed common acceleration “a” are shown in the figure. Here all forces are along y-direction only. Now,

3 g - k x 0 = 3 a 3 g - k x 0 = 3 a

and

k x 0 - 2 g = 2 a k x 0 - 2 g = 2 a

Combining two equations, we have :

a = g 5 a = g 5

Putting in first force equation, we have :

3 g - k x 0 = 3 a = 3 x g 5 k x 0 = 3 g - 3 g 5 = 12 g 5 3 g - k x 0 = 3 a = 3 x g 5 k x 0 = 3 g - 3 g 5 = 12 g 5

For x 0 x 0 = 0.01 m,

k = 12 g 5 x 0 = 12 x 10 5 x 0.01 = 2400 N / m k = 12 g 5 x 0 = 12 x 10 5 x 0.01 = 2400 N / m

### Variable acceleration

Spring, however, behaves differently for a certain period in the beginning. When we apply force, spring begins to stretch or compress. During this period, the spring force is not constant but its magnitude depends on the extension as :

F = -kx

where "x" is the displacement. Clearly, spring force is opposite in direction to the displacement. As force is variable here, acceleration of the body attached to spring is also variable. We can not handle variable acceleration, using equation of motion for constant acceleration. The detail of spring motion under variable acceleration, therefore, is handled or analyzed using energy concepts.

We can, therefore, analyze spring motion only in limited manner during which length of spring changes. For illustration, consider the case of two blocks, which are connected by a spring and pulled by an external force.

The accelerations of the blocks are not same in this period. It must be understood that extension of the spring takes finite time to extend or compress during which blocks will have different accelerations and spring will be subjected to different forces at the two ends. Eventually, however, extension is stabilized for the given external force and the blocks move with same accelerations.

#### Example 4

Problem : Consider the configuration as given in the figure. If “a” be the acceleration of block “A” at the instant shown, then find the acceleration of block “B”.

Solution : It is important to note the wording of the question. It assumes accelerations to be different. It means that the spring is getting extended during the motion. We do not know the spring force.

Let us consider that spring force be F s F s .

Free body diagram of “A” Free body diagram of “A”

F x = F s = m a a = F s m F x = F s = m a a = F s m

Free body diagram of “B” Free body diagram of “B”

F x = F - F s = m a 1 m a 1 = F - F s F x = F - F s = m a 1 m a 1 = F - F s

Combining two equations and eliminating, F s F s , we have:

a 1 = F m - a a 1 = F m - a

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