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Tension in ropes between accelerating bodies

  • Thread starter quietbang
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Hi, I am currently enrolled in a first year Physics for Scientists and Engineers course at a university, but I am a biochemistry student without a physics background. I am relatively certain that the problem is a forces problem, but there is a chance that it could be energy as we have just begun studying that. I am not sure how to calculate tensions over a non-constant system, particularly if that system is accelerating and not in equilibrium.

Homework Statement



Two 4.56kg blocks connected by a rope are shown.https://loncapa.physics.mcmaster.ca/enc/51/790fe24cb9a36f8b53a9e8421f30c0ad6b5a395de2fa875ae6adf844c8b170def7fe1b61fb49361e0f49fc285eca551db6c419d373e4c41d287a8141ea65c13e.png A second rope hangs beneath the lower block. Both ropes have a mass of 253.0g. The entire assembly is accelerated upward at 3.55m/s2 by force F.

a)What is force F?

b)What is the tension at the top end of rope 1?

c)What is the tension at the bottom end of rope 1?

d)What is the tension at the top end of rope 2?

Homework Equations


[itex]
\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}[/itex]

The Attempt at a Solution


I solved for F by using [itex]\vec{F}_{net_y} = \Sigma \vec{F}_{y_total} = m_{total} \vec{a}_{total}[/itex]

the acceleration of the system is additive, therefore it is

[itex]a= 9.81m/s^2 + 3.55m/s^2[/itex]

[itex]a=13.36m/s^2[/itex]

Total mass of the system is 9.63Kg.

[itex] \vec{F}_{net y} = 9.63kg*13.36m/s^2[/itex]

[itex]F=129N[/itex]

This was the correct answer.

For the next part of the problem, I tried to solve for the bottom of mass A, as that tension would be the same or similar to the tension at the top of the Rope 1.

To do so, I calculated that

[itex] \vec{F}_{net y} = \Sigma \vec{F}_{y} = m_{total} \vec{a}_{total}[/itex]

[itex]\Sigma \vec{F}_{y} = F-T= m_{total} * \vec{a}_{total}[/itex]

[itex]129N-T=m_{total} * \vec{a}_{total}[/itex]

[itex]129N-T=4.56kg*(3.55*9.81)[/itex]

[itex]T=129N*60.9N[/itex]

[itex]T=2.12N[/itex]

This is not the correct answer. I also tried doing it without adding the accelerations and using only the applied acceleration on the system, and got 7.96 N as a result. This is also wrong.
I can't work any further on the problem, as I need this result to calculate the others.
 

Answers and Replies

  • #2
Simon Bridge
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Two 4.56kg blocks connected by a rope are shown.https://loncapa.physics.mcmaster.ca/enc/51/790fe24cb9a36f8b53a9e8421f30c0ad6b5a395de2fa875ae6adf844c8b170def7fe1b61fb49361e0f49fc285eca551db6c419d373e4c41d287a8141ea65c13e.png A second rope hangs beneath the lower block. Both ropes have a mass of 253.0g. The entire assembly is accelerated upward at 3.55m/s2 by force F.
The image is password protected.

I solved for F by using [itex]\vec{F}_{net_y} = \Sigma \vec{F}_{y_total} = m_{total} \vec{a}_{total}[/itex]
I would have expected you to use free-body diagrams.

the acceleration of the system is additive,
I don't know what that means?

[itex]a= 9.81m/s^2 + 3.55m/s^2[/itex]
You mean that the whole system accelerates faster than gravitational acceleration - so that the force must be acting in the same direction as gravity?
 
  • #3
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The image is password protected.
Sorry, I've attached it this time.

I would have expected you to use free-body diagrams.
I did initially, but the force answer I got was way off.

Separating the masses into two diagrams, I was left with these equations:
There aren't any forces in the x-direction, so their force equations are 0=0.
For the y direction, mass A
[itex]\vec{F}_{y}= \vec{F}_{1}-W_{A}-\vec{N}_{rope 1 on box a} = ma[/itex]
(equation 1)

For the y direction, mass B
[itex]\vec{F}_{yB}=\vec{T}-\vec{W}_{B}-\vec{N}_{rope 2 on box b} =m_2 a [/itex]
(equation 2)

Using equation two, [itex]\vec{T}=m_2 a + m_2 g +\vec{N}_{rope 2 on box b}[/itex]

[itex]\vec{N}_{rope 2 on box b}=[(4.56kg*3.55m/s^2)+(4.56kg*9n8m/s^2)+(.253kg*9.8m/s^2)][/itex]

[itex]\vec{N}_{rope 2 on box b}=63.35 N [/itex]

Subbing this value into mass 1, by Newton's Third Law

[itex]\vec{F}_{1}=M_{a}G - \vec{N}_{rope 2 on box b} =m_{A} a[/itex]

[itex]\vec{F}_{1}= m_{A} a + M_{a}G + \vec{N}_{rope 2 on box b}[/itex]

[itex]\vec{F}_{1}= [ (4.56kg * 3.55m/s^2) + (4.56 * 9.8m/s^2) + 63.35N ][/itex]

[itex]\vec{F}_{1}= 124.23N[/itex]
This is incorrect.

I then looked in my textbook, where it said that because the force has to overcome gravity and accelerate the system, when using f=ma you need to take gravity into account, and the acceleration is the actual acceleration of the system plus the force necessary to overcome gravity in order to travel in an upwards direction. Thus, you can calculate the force necessary to move the whole system at an acceleration of 3.55m/s^2 by adding the masses of the system, and adding the force of gravity to the acceleration, as it must be overcome. You then solve for F. I did this, and was told that my answer was correct. This may not be the best method to use, but it was the only one I could find that accounted for gravity.
 

Attachments

  • #4
Simon Bridge
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Separating the masses into two diagrams [...]
The ropes have mass too.
 
  • #5
arildno
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Note what Simon Bridge tells you: The ropes have mass.

What does that entail?

You should clarify for yourself WHY, for the massless (m=0) rope, we have that the tension is constant throughout the rope!

IF the forces acting on a massless rope segment at each end (i.e, the tensions) had been imbalanced, so that the net force F on the rope segment had been non-zero, then Newton's second law for the rope segment would read a=F/0=infinity!!

Therefore, if the massless rope segment is to have a finite acceleration, the tensions at each end of it must be equal, so that F=0
-----------------------------------------
For a rope with mass, the tensions at each end must typically be different, in order to generate acceleration in the rope segment
 

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