String tension

[georgeo]

I have a hard time conceptualizing string tension, and hence calculating it.

For a hypothetical situation, there is a massless string connecting two blocks on a flat, frictionless surface. Block A is 5 kg, block B is 3 kg. There is a string attached to the outside of Block A, and a force of 150 N is applied to this string.(so the outside string pulls block A, which pulls the middle string, which pulls Block B)

This is not a homework problem that i know of at all, I just need to conceptualize and calculate tension in a simple situation, and then I should be able to apply to other situations too.

Thank you everyone!

 

[Pyrrhus]

Well imagine a string tied someplace, when you pull it you exert a force to the object where the string is tied, and at the same time the string exerts a force on you.

 

[georgeo]

Yeah, I get that part.

So i guess the question is, how do I calculate the tension?

 

[Pyrrhus]

Draw a Free Body Diagram

I suppose the system is at rest (V = 0)

Use Newton's 1st Law

$$\sum_{i=1}^{n} \vec{F}_{i} = 0 \rightarrow \vec{V} = constant$$

 

[georgeo]

Lets say there is a massless string hanging from the ceiling and on the end is a 1kg mass. The tension would be 9.8 N, right?
What if the string goes over a pulley, and on the other side is a 2 kg mass? The system would accelerate. What would the tension of the string be?

 

[Gokul43201]

The tension in the string will be the same throughout it. So looking at each of the 2 blocks, write down the forces on each block (weight = mg downwards; T upwards). The net force causes an acceleration a, given by Newton's Third Law. Writing this equation for the 2 blocks, you have 2 equations, in 2 unknowns, T and a. Solve simultaneously to find them.

 

[georgeo]

ahhh...
i think that's what i need...
i'll let you know how my test today goes :)

(that was just the way of saying it that made it click in my head)

 

[georgeo]

I got an A on my test. Thank you all!

 

[Pyrrhus]

I got an A on my test. Thank you all!

Excellent! Congragulations!

 

[skylion80]

I have a similar problem. There is a 60 kg block on a frictionless table. At the corner of the table there is a pulley. A string is attached to the block and runs over the pulley, where it is attached to a 25 kg block that hangs off the table. It looks something like this:

````____
___|_60_|-------------O
############$#| |$##########$#| |$##########$#| _|_$##########$#||25|$#############|

I am supposed to calculate the tension on the string and the acceleration of the 25 kg block, but the formulas I have tried don't work. Can anyone help?

PS Please keep the formulas simple. I'm not good at physics, and this is my first class.

 

[Ken G]

The key point is that a massless string cannot support any net forces along its length-- it would have to have infinite acceleration! So the rule about such a string is that the whole stretched string has the same velocity and acceleration at all times, and whatever force pulling one way on one end equals the force pulling the other way on the other end. Once you've put that into your calculation, you are done with the string, and you need to concentrate on the masses. You apply F=ma to the masses, given those rules about the string. Often, as above, this will mean asserting the force at both ends of the string is the tension "T", but not knowing what T is. That's fine, since you write F=ma for each mass, all you need is that T is the same in both equations, and you can do the problem. You are looking for the same T and the same a involved in the F=ma for both masses, and so you can solve for both of them simultaneously-- only one combination of T and a will satisfy F=ma for both masses (where F is all the forces on the masses-- the T and the gravity and anything else too). In your problem, it helps to concentrate on the F=ma "along the string direction", and ignore what is happening in other directions. Do you know that the T exerted on the masses at both ends of a string must point in toward the middle of the string? (Strings only pull, they don't push.)