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In my textbook, the definition of tension was given that Tension is the reactive force which exists when string is being stretched at its both end. After it there was a case given that to calculate acceleration and tension produced in string while it's attached with different masses at both its ends and passed over a frictionless pulley. First body is being hung and second is placed over horizontal frictionless surface with relatively smaller mass. It is given in my book that there are three forces acting on the body placed on horizontal smooth surface viz., Weight of body, the reaction force to weight and tension in string pulling body towards pulley's wheel. How can there be tension in string according to the definition given my book however there is no acting force (pulling force) on the body that pulls it away from wheel.

There are different definitions for tension. The one in our book is creating problem however second one, that is, " tension describes the pulling force exerted by each end of a string, cable, chain, or similar one-dimensional continuous object " is al right.

Since acceleration is lowered due to Tension then acceleration produced in the body must be given by, $a=\frac{Net-Force}{m}$ $=>$ $a=\frac{m_1g-T}{m_1}$ since $W_1$ $>$ $T$.

If tension in string is the reactive force to the weights pulling the string apart then according to Sir Isaac Newton, $W_1=W_2=T$.

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  • $\begingroup$ I've seen definitions that vary slightly, but if that is what you text says, then you should probably be ready to use that version in class and on assignments... $\endgroup$ – dmckee Feb 22 '15 at 21:30
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    $\begingroup$ I want the right concept. I don't care what my teacher and textbook says. $\endgroup$ – user66452 Feb 22 '15 at 21:31
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    $\begingroup$ There are at least two ways to formulate a definition for "tension" that generate correct physics. They are both "right". You can define tension as the pair of forces trying to pull the string apart or as the strings reaction to those forces. Because of Newton's 3rd law both are generate the same results. The thing is that if you need to explain yourself, you should use the version that your instructor is expecting. $\endgroup$ – dmckee Feb 22 '15 at 23:05
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    $\begingroup$ Here's a third: tension is the force that a string (or any other object that can "pull") applies to an object. That object may be another segment of the same string. (i.e., choose a point in the string. At that point, the left half applies tension to the right half, and vice versa.) This one doesn't need the potentially confusing use of two forces, and it avoids the introduction of reaction force, which sounds to some like backwards reasoning. Like the others, this works correctly if carefully applied. (But you will likely have to understand the definition in your book.) $\endgroup$ – garyp Feb 23 '15 at 1:08
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As the weight hanging off the table falls, it has some acceleration downwards. The force on the body is entirely due to gravity and is known. The string acts as a linkage between the hanging mass and the mass on the table. Really in this case, we have weights $m_1$ and $m_2$ being pulled by a force $m_1g$, which leads to the acceleration of the system being lower than a free-falling weight.

We can also use the concept of tension as a reactionary force to draw in the forces. The hanging weight $m_1$ is falling with some acceleration less than a free fall. However, the force of gravity on the falling weight is definitely still $m_1g$. There must be another force acting on the object pulling it upward to counter the pull of gravity. This is the tension $T$ in the string. The string pulls the weight up with enough force mitigate the pull of gravity and slow the acceleration. This tension can be thought of as going up through the pulley to the weight on the table and causes the table weight to start sliding. Then one can say the only horizontal force on the table weight is the $T$ from the string.

Tension is a reactive force by it's nature. It's the same as the ground. When you step on the ground, the ground 'pushes back' with an equal force. When you pull on the string, it 'pulls back' equally.

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  • $\begingroup$ I think it's the first part that's confusing you, when I say the force m1g is pulling mass m1+m2. The reason I say this is the string essentially just acts as a link between the two blocks. If both blocks were both on a horizontal surface, you could shrink the string between them until they were the same block and the behavior of the system wouldn't change. That's an equally valid way of looking at the situation as the tension picture. Tension is more useful in real life though where strings or pulleys can break. $\endgroup$ – MonkeysUncle Feb 23 '15 at 13:07
  • $\begingroup$ Looks like you are considering string off the system. When you do this, tension is out of the context however I asked if tension is the reactive force there. Well, if you want to consider tension as an opposing force to the motion, you have to define where it comes from. And, you have to explain how it goes up through the pulley and pulls the body on horizontal. $\endgroup$ – user66452 Oct 30 '15 at 4:35
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    $\begingroup$ "Tension is a reactive force by its nature...it pulls back 'equally". I think this is extremely silly. I asked how tension is the reactive force and you answered tension is the reactive force. $\endgroup$ – user66452 Oct 30 '15 at 4:39
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The best way to define tension is as the one dimensional version of the stress tensor. So, you can define the tension at some point P in the string as the force at which the part of the string on one side of P pulls on the other side of P. The direction of this force then depends on which sides you are considering, so you should make some choice here and relative to that choice the tension is well defined.

Suppose that a sting hangs over a pully, as a result the direction of tension changes. But how do we explain that the magnitude of the tension is the same on both sides? If you are stuck with defining tension as a reaction force to whatever is pulling on it from its ends, then you'll not be able to even address this question. Most likely you'll just assume it's the case without really understanding why.

With the proper definition, you can do a balances of forces calculation along the string at the points where it makes contact with the pulley. If you do this, you'll see that it boils down to the fact that the pully exerts a force perpedicular to the string, this has the effect of changing the direction of the tension, but not its magnitude.

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