| << Chapter < Page | Chapter >> Page > |
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.
Balanced force system
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
Mass of the string below point A, " ", is :
The external forces at point "A" are (i) Tension, T (ii) weight of block, mg, and (iii) weight of string below A," ".
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.
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 :
where 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.
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,
Spring and pulley system
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,
Free body diagram
and
Combining two equations, we have :
Putting in first force equation, we have :
For = 0.01 m,
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.
Block spring system
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.
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”.
Block spring system
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 .
Free body diagram
Combining two equations and eliminating, , we have:
Notification Switch
Would you like to follow the 'Physics for k-12' conversation and receive update notifications?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|