Can't understand 2 pulley 3 masses problem

In summary: I'm still having a hard time understanding how to solve this problem. I've tried to do it step by step but I'm still not getting it. I've tried to find the equations for both the x- and y-components of the tension forces, but I'm not sure if it's correct to do that. Any help would be greatly appreciated!The masses of the three objects at the other end of the ropes are equal to double their equivalent in Newtons.
  • #1
TissueCulture
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0

Homework Statement


System is in equilibrium. 1 mass known: 79.4 kg (or 778.6 N) and 2 other unknown. Other masses unaccounted for. Everything friction-less. The Tensions are unknown also.

Homework Equations


∑F = 0
∑Fx = -T1*cos135 + -T3*cos37 = 0
∑Fy = (-T1*sin135 - m1g) + (-T3*sin37 - m3g) + T2 - 778.6 N = 0

The Attempt at a Solution


I don't have any idea how to comprehend this problem. Of course I am missing something but I don't know how to ''think'' the problem from the get go. Any pointers would be very appreciated, I have tried to figure it out for the last hours...The T1 and T3 arrows should be parallel to the rope on the drawing
1epe8x.jpg
 
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  • #2
The key is to consider the forces acting on the red knot where the three strings meet. The knot is in static equilibrium. Below is a figure of the knot and the three strings. Use that as the beginning of a free-body diagram of the knot. You should draw the force vectors acting on the knot and label them. Be sure to put arrow heads on the force vectors to indicate the directions of the forces.
 

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  • #3
So if I understand correctly, you mean that I should consider that T1,T2 and T3 vectors are pulling away from the red knot instead of towards it? Wouldn't that make them go in the same direction as the m1g, m2g and m3g vectors? Or is it an error on my part to consider the masses of the objects as vectors instead of mere units of mass? Thank you very much for your help!
 
  • #4
Yes, the three tension forces acting on the knot all point away from the knot along the strings. There are no other forces acting on the knot.
 
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  • #5
One thing I can't get my head around is if I must substract the masses with both ∑Fx and ∑Fy

∑Fx = -T1*cos135 - m1g + T3*cos37 + m3g = 0
∑Fy = (T1*sin135 + m1g) + (T3*sin37 + m3g) - T2 - 778.6 N = 0

Are these valid to start the problem? I have a hard time understanding how to ''translate'' the force vectors of the free-body diagram into mathematically sound equations for both X and Y axis. Thanks for your help anyone!
 
  • #6
There are only three forces acting on the knot: T1, T2, and T3. So when you write out ∑Fx and ∑Fy, you should only use these three forces.

When you write the x-component of T1, you should write it as T1cos135o and not as -T1cos135o. If you calculate cos135o, you will see that it is a negative number. So, T1cos135o is already negative. You shouldn't add an extra negative sign in front.

However, if they had specified the angles as shown below, then the x-component of T1 would be written with a negative sign as
-T1cos45o
 

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  • #7
Thank you very much! in fact, both angles had to be measured in real life but I thought I had to consider the angle in relation to the positive X axis (I don't know if this is clear), so I ended up with 135 degrees. However, I suppose it is more logical or easier to go with 45 degrees now that you told me this...! What got me bugged is how I am including the weights in the equations (m1g, m2g (778.6N) and m3g). Does it seem like I'm on the right track?
 
  • #8
Try to write out ∑Fx = 0 and ∑Fy = 0 for the knot. The only forces that should appear are the x or y components of the three tension forces.

However, you can then proceed to relate the tension forces to the weights of the three objects at the other end of the ropes. To do this, construct a separate free body diagram for each of the three objects and apply the condition for equilibrium.

You can assume that the tension at one end of a rope is equal to the tension at the other end of that rope.
 
  • #9
Hi again! Here is what I found so far, if correct I only have one other question!

∑F = T1 + T2 + T3 (with vector arrows)
∑Fx = -T1cos45 + T3cos37 = 0
∑Fy = T1sin45 - T2 + T3sin37 = 0

1) -T1cos45 + T3cos37 = 0
T1 = (T3cos37)/cos45 = 1,129T3

2) T3(1,129*sin45 + sin37) - T2 = 0
T3(0,7986 + sin37) - 778,6N = 0
T3(1,400) = 778,6N
T3 = 778,6N / 1,400 = 556,1 N

3) T1 = 1,129 * 556,1N = 627,8 N

If the above is OK, I just wonder if the masses for the objects are equal to double their equivalent in Newtons, so: m1 = 2*T1 (in N) ??

Thanks a lot for your help TSny! :)
 
  • #10
TissueCulture said:
Hi again! Here is what I found so far, if correct I only have one other question!

∑F = T1 + T2 + T3 (with vector arrows)
∑Fx = -T1cos45 + T3cos37 = 0
∑Fy = T1sin45 - T2 + T3sin37 = 0

1) -T1cos45 + T3cos37 = 0
T1 = (T3cos37)/cos45 = 1,129T3

2) T3(1,129*sin45 + sin37) - T2 = 0
T3(0,7986 + sin37) - 778,6N = 0
T3(1,400) = 778,6N
T3 = 778,6N / 1,400 = 556,1 N

3) T1 = 1,129 * 556,1N = 627,8 N

Your results for the tensions look correct and your work is nicely written out.

If the above is OK, I just wonder if the masses for the objects are equal to double their equivalent in Newtons, so: m1 = 2*T1 (in N) ??
I'm not sure why you would think that. Suppose you want to find m1. Draw a free body diagram for the object with mass m1. How many forces act on this mass? What direction do they point? Is m1 in equilibrium? Do you know the value of any of the forces acting on m1?
 
  • #11
What I mean to say is, I thought that a pulley divided by a factor of 2 the force necessary to lift it. I'm not sure exactly where I have read that, I cannot find it in my notes...Maybe I am mistaken! I sure know now that T1 is 627,8N but my brain seems to tell me there is more to it than simply T1 = m1 (which needs to be converted into kg). Am I just crazy?

Thanks for the encouragement by the way!
 
  • #12
Aaaah I think I get why I thought it divides the force by 2. I somehow subconsciously considered this system as a combined pulley system, with each pulley being fixed of course but the ''knot'' acting as a movable pulley, which is not the case!

I now assume that it is clearly T1 = m1, since :
'' There is a force (tension) on the rope that is equal to the weight of the object.
This force or tension is the same all along the rope. In order for the weight
and pulley (the system) to remain in equilibrium, the person holding the end
of the rope must pull down with a force that is equal in magnitude to the
tension in the rope. For this pulley system, the force is equal to the weight, as
shown in the picture. The mechanical advantage of this system is 1'' I cannot post the picture but it shows a simple pulley with a weight attached. Found this online.
 
  • #13
Yes. However, keep in mind that mass is not the same as weight. So, you would not say that T1 = m1. Otherwise, I think your reasoning is good!
 
  • #14
Yes I was about to add that in fact, the force being already known, it would rather go like: T1 = m1*g → M1 = T1/g

Which gives the mass!

Thanks a lot again for your help!
 
  • #15
TissueCulture said:
Yes I was about to add that in fact, the force being already known, it would rather go like: T1 = m1*g → M1 = T1/g

Which gives the mass!
Yes, good work.
 

Related to Can't understand 2 pulley 3 masses problem

1. How do you determine the tension in the ropes for a 2 pulley 3 masses problem?

In order to determine the tension in the ropes for a 2 pulley 3 masses problem, you will need to apply the principles of Newton's laws of motion. This involves setting up a system of equations and solving for the tensions in the ropes using the known masses and accelerations.

2. Why is a 2 pulley 3 masses problem considered more complex than a single pulley problem?

A 2 pulley 3 masses problem is considered more complex because it involves multiple masses and multiple ropes, making it harder to determine the tensions and accelerations. In a single pulley problem, there is only one mass and one rope, making it easier to solve.

3. Can you use the same approach for solving a 2 pulley 3 masses problem as you would for a single pulley problem?

Yes, you can use the same approach for solving a 2 pulley 3 masses problem as you would for a single pulley problem. However, you will need to take into account the additional ropes and masses in the 2 pulley 3 masses problem, which may require more complex equations and calculations.

4. How do you draw a free body diagram for a 2 pulley 3 masses problem?

To draw a free body diagram for a 2 pulley 3 masses problem, you will need to identify all the forces acting on each mass, including the tension in the ropes, the weight of the masses, and any external forces. Then, you can draw arrows to represent the magnitude and direction of each force.

5. Are there any real-life applications for a 2 pulley 3 masses problem?

Yes, there are many real-life applications for a 2 pulley 3 masses problem. This type of problem can be found in various engineering and mechanical systems, such as cranes, elevators, and pulley systems used in construction. It is also commonly used in physics and mechanics courses to demonstrate the principles of Newton's laws of motion.

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