Does Using a Pulley Affect Acceleration in a Frictionless System?

AI Thread Summary
In a frictionless system with a mass connected to a string over a pulley, the acceleration of the mass on the surface is affected by the weight of the second mass. The tension in the rope, caused by the weight of the second mass, pulls up on the first block, decreasing its net force. This results in a lower acceleration than predicted by the equation a=g(m). The pulley itself does not contribute additional forces but changes the direction of the tension force. Understanding this concept can be aided by drawing a free-body diagram to visualize the forces at play.
just.karl
Messages
64
Reaction score
0
With a mass on a frictionless surface connected to a string that is sent over and down to a mass. Would that increase or decrease the acceleration compared to the equation, a=g(m). I'm just stumped on how to think of this. The pulley is not included in the equation but how would I figure out the difference?
 
Physics news on Phys.org
The extra mass attached to the rope will cause a tension in the rope correct? How does this tension affect the net force on the original block?
 
It would decrease the net force on the original block, correct? But would it be just because of friction or is it something more complicated than that?
 
Yes, the net force would decrease but not for the reason you are suggesting.

Consider the pulley to be frictionless. It's not the pulley that is decreasing the force on the block, but the weight of the second block that is added. The weight of the block causes a tension in the rope, and the rope thus pulls up on the first block, decreasing the net force on the first block. All the pulley really does is change the direction of the tension force so it pulls up on the block. This is why the pulley doesn't factor into the force equation, since it doesn't create any new forces.

(Of course, in reality, there is some friction in the pulley which will decrease the net force on the block, but it is not the primary cause of the decrease.)

Does this make sense? I suggest drawing a force/free-body diagram of the situation. It may help you visualize what's happening.
 
make it as simple as possible first

just.karl said:
With a mass on a frictionless surface connected to a string that is sent over and down to a mass. Would that increase or decrease the acceleration compared to the equation, a=g(m).

I assume by "frictionless surface", you mean a horizontal one, like a really smooth table?

Try an easier problem first: tilt the table, and put the pulley on the end of the table, so that the string always stays in the same direction.

Can you solve that problem?
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top