Tension in a rope on frictionless pulley

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SUMMARY

The discussion revolves around calculating the tension in a rope connected to a 4kg object on a table and an 8kg weight hanging off a frictionless pulley. Key equations utilized include Newton's second law (F=ma) and the frictional force equation (F of friction = coefficient of friction × Force normal). The participants emphasize applying these equations separately to each mass and combining the results to solve for the unknown tension. The final approach involves setting up two equations based on the forces acting on each mass and solving for tension.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of frictional forces and coefficients
  • Ability to set up and solve simultaneous equations
  • Familiarity with basic mechanics involving pulleys
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  • Learn how to derive equations of motion for systems involving pulleys
  • Study the concept of tension in strings and ropes in physics
  • Explore examples of frictionless pulley problems in mechanics
  • Practice solving simultaneous equations in physics contexts
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Students preparing for physics exams, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to tension and friction in pulley systems.

simon360
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Homework Statement



A 4kg object is palced on a table with a coefficient of friction of 0.2, and attached to a rope. This rope is run through a frictionless pulley, and attached to an 8kg weight. What is the tension in the rope?

Homework Equations



F=ma
F of friction = (coefficient of friction)(Force normal)

The Attempt at a Solution



Force normal of weight = 8(9.8) = 78.4
Force normal of object = 4(9.8) = 39.2
Force of friction on object = 0.2(39.2)

I don't know where to go from here, and I have an exam tomorrow. Help would be greatly appreciated, obviously :P

Thanks, I hear you guys are the best for quick physics help ;)
 
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Apply Newton's 2nd law to each mass, then combine the two equations.
 
So add both forces from F=ma together?
 
simon360 said:
So add both forces from F=ma together?
Not sure what you mean. What I mean is: Apply F=ma to each mass separately. You'll end up with two equations that you will solve together.

What forces act on each mass?
 
Doc Al said:
Not sure what you mean. What I mean is: Apply F=ma to each mass separately. You'll end up with two equations that you will solve together.

What forces act on each mass?

Well, the weight is pulling the object because it is hanging off the table, through the pulley...
 
What pulls on the object is the tension in the string. (Of course that tension also pulls up on the weight.)
 
Ahh, I think I get it (but physics has confused me from the start, so bear with me). The tension is essentially a net force?
 
No. "Net force" just means the total force on an object. The tension is just one of several forces.
 
Alright, I'll be honest, I still don't get it :(

This was a quick question I came up with to demonstrate what I need to know. What would be the tension? Then I could figure out how to actually get it.
 
  • #10
Start by applying Newton's 2nd law to the hanging mass. What forces act on it?
 
  • #11
Gravity times its own weight, and the object on the table (which I presume can act as friction). Anything I'm missing?

The object on the table is also prone to friction, and the weight that is hanging pulls back on it.
 
  • #12
simon360 said:
Gravity times its own weight, and the object on the table (which I presume can act as friction). Anything I'm missing?
Stick to the forces acting on the hanging mass. Yes, its weight is one force. What's the other? (Forget about the object on the table for the moment. We'll analyze that one next.)
 
  • #13
I can't think of any, unless you mean the rope/tension itself.
 
  • #14
simon360 said:
I can't think of any, unless you mean the rope/tension itself.
Of course I mean the rope tension!

There are two forces acting on the hanging mass: The rope tension, which acts up; the weight, which acts down.

Now apply Newton's 2nd law.
 
  • #15
Haha, thought that would be too easy :P

The force down = 8(9.8) = 78.4 N

Not sure how to deal with the rope tension, tbh...
 
  • #16
The rope tension is unknown, so just call it T. (You'll end up solving for it.)
 
  • #17
Oh, ok.

So the net force is F down - T
 
  • #18
The net force on the hanging mass is T - m_2 g, where m_2 equals 8 kg.
 
  • #19
Alright, so essentially just the reverse: T - 78.4.
 
  • #20
OK, now apply Newton's 2nd law.
 
  • #21
I thought I already did...

Would I be right to say that it's 7.84 = T - 78.4?

Just because the coefficient of friction is 0.2, force normal on the object is 39.2, and thus friction is 7.84.
 
Last edited:
  • #22
Apply Newton's 2nd law to the hanging mass only. Friction has nothing to do with it. Call the acceleration "a".
 
  • #23
Well, mass * a = 8 * 9.8 = 78.4. I really don't understand what you want me to do after that...
 
  • #24
simon360 said:
Well, mass * a = 8 * 9.8 = 78.4. I really don't understand what you want me to do after that...
The acceleration is not 9.8 m/s^2. That's the acceleration of a freely falling object, not something attached to a string. Call the acceleration "a".
 
  • #25
I truly and honestly don't know what to do.
 
  • #26
Newton's 2nd law:
F_{net} = m a

T - m_2 g = m_2(-a) = -m_2a

I called the magnitude of the acceleration "a". Since it accelerates downward, the acceleration = -a. (Up is positive, down is negative.)

See if you can apply Newton's 2nd law in a similar manner to the other mass.
 
  • #27
I guess the * things mean multiply, so you wrote mg=ma... but the net force isn't mg, it's mg-T I think
 
  • #28
Ahh, that makes more sense...

I think this is the only thing I will have trouble with on the exam, and I really have to go to bed since it's 11:40 here. Thanks for your help, but I think sleep is more important than the few potential marks I'll lose here...
 
  • #29
wait, might as well tell you the answer first...
 
  • #30
first equation, m1g-T=m1a
second equation= T-um2g=m2a

by u I mean coefficient of frictions, um2g is just the Ff=uFn equation

notice that the acceleration is the same for both things so solve for acceleration in one equation and substitute that into the a in the other equation
 

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