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My understanding of tension..by holezch
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#1
Jan2610, 04:00 PM

P: 253

Hi, according to my understanding of tension, if I had two blocks of equal masses connected by a cord and a pulley (one on a cliff and the other hanging off a cliff), would they be in equilibrium? The block hanging would pull the cord down by mg, and then the cord will pull the block on top, and then by newton's third law, the block will pull back equally.. Is this right? thank you



#2
Jan2610, 04:14 PM

P: 230




#3
Jan2610, 04:23 PM

P: 253

thanks for the reply, I'm assuming that there is no friction. Could you tell me what went wrong with my "analysis"?
mg pulls on cord from the bottom, cord pulls on top mass by mg but by newton's third, it pulls back on the cord by mg? So the force applied to the top mass would've been mg but it's mg  mg because of the reaction force..? thanks 


#4
Jan2610, 04:55 PM

P: 269

My understanding of tension..
I don't understand where the pulley comes into this system. Please elaborate.



#5
Jan2610, 05:02 PM

P: 253

http://www.physics247.com/physicshomeworkhelp/008.jpg
I found a picture of something that is visually like my scenario, only in my case, all the masses are equal thanks 


#6
Jan2610, 05:10 PM

P: 269

Ahh I understand now, and yes, without friction and assuming the pulley is ideal, the block on the cliff would slide off. This is because what the block hanging off the cliff is doing is applying a tension force to the rope, which is causing the rope to apply a force on the block on the cliff. Without friction, the forces are unbalanced, which means the block would move and fall off the cliff.
A situation where the two blocks would be in equilibrium is if the cliff was removed, and both blocks were hanging; with the pulley in the middle of the rope. Hope this helps. 


#7
Jan2610, 05:16 PM

P: 253

Thanks, yes but wouldn't the block on the cliff pull back on the cord with the same force?



#8
Jan2610, 05:52 PM

P: 253

oops, I don't mean equilibrium I guess, I mean that the hanging block won't fall down
thanks 


#9
Jan2610, 07:02 PM

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P: 12,069

The mass on top has a net force acting to the right, due to the tension in the rope. With no friction, that is the only horizontal force acting on that mass, therefore it must accelerate to the right. 


#10
Jan2610, 07:23 PM

P: 269

So, in a frictionless situation, the block will fall. In situation with friction, the block might or might not fall; depending on the coefficient of static friction. http://upload.wikimedia.org/wikipedi...iction_alt.svg This is a pretty good picture of a free body diagram of something similar to your problem. Just imagine that your rope is tied to the left side of the block, that contact would give you your 'F' force, the force that is trying to move the block. The friction force then is parallel to the surface the block is placed on, and resists movement. That is the [tex]F_{f}[/tex] force. There is no internal force of the block that resists movement, it is friction. 


#11
Jan2610, 07:36 PM

P: 253

thanks for the replies guys, RedBelly89 , I guess I wasn't talking about the top block, I'm really asking if the dangling block will move down? (I don't think so, since by newton's third law, the top block pulls back on the rope by the same force)
and KrisOhn, I was talking about the reaction force, by newton's third law, as the cord pulls the top block, the top block will pull the cord.. so will the dangling block move? it seems impossible that the dangling block won't move as the top one does.. but applying newton's laws seems to suggest this to me.. thanks guys 


#12
Jan2610, 07:47 PM

P: 269

And yes, the reaction force from the dangling block`s weight is essentially force `F` in the diagram I linked to. And the `top block pulling the cord` force is friction. As for whether the block moves or not, I cannot say, it is dependent on the coefficient of static friction for that surface. I`m not sure how to explain it much simpler than that. 


#13
Jan2610, 07:57 PM

P: 253

But the top block will pull the cord even if there is no friction, isn't that true?? it exerts a reaction force on the cord? the cord pulls down the block, so the block pulls the cord up. :S 


#14
Jan2610, 08:07 PM

P: 269

I'm going to use an analogy, explaining this should not be so difficult. Let's say you're skiing, you tie a rope to a rock that is the same mass as you, and then kick the rock off the side of a cliff. You hold onto the rope. When the rope becomes taught, there is a force exerted on your hands by the rope, this is similar to the force exerted by the rope on the block. And since you are on skis, the coefficient of static friction is low enough that you will start to slide, and acceleration will occur. 


#15
Jan2610, 08:13 PM

P: 253

thanks 


#16
Jan2610, 08:22 PM

P: 269




#17
Jan2610, 08:27 PM

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You're assuming that the whole system is in free fall. It's not. Consider the system of two masses. The net force on the system is just the weight of the hanging block (mg). But the total mass of the system is 2m. So the acceleration of the system is g/2. Now, with what force much *each* block be pulled on in order to accelerate at g/2? The answer is that each block has a net force of mg/2 on it. Since the hanging block has a force of mg acting on it downward, there must therefore be an upward force of mg/2 acting on it (in order to get the required net downward force). That result tells you what the tension in the rope is. It's totally consistent as well, because if the rope is pulling with mg/2 at one end, then it is pulling with force mg/2 at the other end as well (explaining the acceleration of the block on the surface).
Newton's Third Law is certainly true. The force with which the hanging block pulls down on the rope is certainly the same as the force with which the the rope pulls up on the hanging block. 


#18
Jan2610, 08:38 PM

P: 253

EDIT: ah yeah, that does make sense, in the free fall version, there are 2 forces moving the system, the weight of the bottom one and the weight of the top one, so the total force of the system is 2mg.. then it's analogous to what you just explained just now.. :) thanks, I realy appreciate it. 


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