Tension in string pulling two carts

In summary, the pulley system does not have the same tension on each side because of the mass in between.
  • #1
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Homework Statement
Figure attached shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force (820t) acting for a brief period of time accelerates the system from rest and acts for 3.00 s. Acceleration is 0.7805t. a) What is the horizontal force acting on the connecting cable between cart A and B at this instant? Assume friction is negligible.
Relevant Equations
F - T = m*a
820t - T = m*0.7805t
T = 820t - m*0.7805t
What would I use as mass? Mass of cart A, or mass of cart A and tractor?
F - T = m*a
820t - T = m*0.7805t
T = 820t - m*0.7805t
I am confused what I would use as mass? Mass of cart A, or mass of cart A and tractor?
 

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  • #2
As with any such question involving interacting masses that can move somewhat independently, draw a separate FBD for each mass. Each FBD should only show forces and accelerations directly related to the mass in question.
 
  • #3
haruspex said:
As with any such question involving interacting masses that can move somewhat independently, draw a separate FBD for each mass. Each FBD should only show forces and accelerations directly related to the mass in question.
cart A's formula: -T + T -Ff = m*a?
 
  • #4
Why have you assumed the two tensions are the same?
 
  • #5
Mister T said:
Why have you assumed the two tensions are the same?
In a pulley system, the tensions are the same, why aren't they the same for this tractor pulling carts?
 
  • #6
annamal said:
In a pulley system, the tensions are the same, why aren't they the same for this tractor pulling carts?
Because of the mass in between. We can only assume tensions are the same both sides of a pulley when there is no friction at the axle and either:
- it is a constant velocity arrangement (maybe static), or
- the pulley has negligible mass

The same applies for linear motion of a body between two strings. In the present case, there is no friction, but the body has mass and is accelerating.

Also, what is Ff in your equation? There is no mention of friction.
 
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  • #7
haruspex said:
Because of the mass in between. We can only assume tensions are the same both sides of a pulley when there is no friction at the axle and either:
- it is a static arrangement, or
- the pulley has negligible mass

Also, what is Ff in your equation? There is no mention of friction.
I think it is negligible
 
  • #8
annamal said:
I think it is negligible
"a) What is the horizontal force acting on the connecting cable between cart A and B at this instant? Assume friction is negligible."

What instant is the problem referring to?
Should we assume it is at the end of the 3 seconds period?

Because the cables linking the three rolling masses do not stretch, the given acceleration is the same for each of those masses.
Cart B can't "see" what forces are beyond the A-B cable; therefore, you can simplify the solution by drawing a free body diagram for B that stops at that cable midpoint.
 
  • #9
annamal said:
In a pulley system, the tensions are the same, why aren't they the same for this tractor pulling carts?
If the pulley has mass, the tensions are not the same. Likewise, for the cart attached to the tractor. Once you master this line of reasoning, it's valuable to think about what happens when the cables are not massless.
 
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  • #10
Mister T said:
If the pulley has mass, and is accelerating, the tensions are not the same.
 
  • #11
Ok, well I solved it until
T = 820t - m*0.7805t. I am just confused which mass I put for m. Mass of cart A or mass of cart A and tractor.
 
  • #12
annamal said:
Ok, well I solved it until
T = 820t - m*0.7805t. I am just confused which mass I put for m. Mass of cart A or mass of cart A and tractor.
That's because you are not using free body diagrams. The diagram will show the forces acting on the object. The vector sum of those forces equals the mass of that object times the acceleration.
 
  • #13
Mister T said:
That's because you are not using free body diagrams. The diagram will show the forces acting on the object. The vector sum of those forces equals the mass of that object times the acceleration.
Free body digram for A:
T2 (tension from string tractor to cart A) <--- A ---> T (tension on string from A to B)
T2 = 820t - m (mass of tractor)*0.7805t
T2 - T = m*a --> T = T2 - m*a = 820t - mT (mass of tractor)*0.7805t = mA (mass of cartA)*0.7805t
Is this correct? I am assuming the tractor and carts are accelerating at the same rate?
 
  • #14
annamal said:
Free body digram for A:
T2 (tension from string tractor to cart A) <--- A ---> T (tension on string from A to B)
T2 - T = m*a
Since this follows from a free body diagram of A, then ##m## is the mass of A.
 
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