Understanding the Force Ratio in Tandem Sleds: Explained with Examples

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The discussion focuses on understanding the force ratio in tandem sleds, where the front sled has twice the mass of the rear sled. The applied force F pulls both sleds on a frictionless surface, with tension T acting between them. The net forces on each sled are analyzed using Newton's second law, leading to the equations T = ma(rear) for the rear sled and (F - T) = 2ma(front) for the front sled. The key point is that both sleds accelerate together, allowing the relationship between their accelerations to be established. The correct force ratio is determined to be 0.33, indicating the tension's magnitude relative to the applied force.
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Two sleds are hooked together in tandem as shown in the figure. The front sled is twice as massive as the rear sled.
The sleds are pulled along a frictionless surface by an applied force F. The tension in the rope between the sleds is T. Determine the ratio of the magnitudes of the two forces,
(a) 0.25
(b) 0.33
(c) 0.50
(d) 0.67
(e) 2.0

Answer is B but why?
 

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gcombina said:
Two sleds are hooked together in tandem as shown in the figure. The front sled is twice as massive as the rear sled.
The sleds are pulled along a frictionless surface by an applied force F. The tension in the rope between the sleds is T. Determine the ratio of the magnitudes of the two forces,
(a) 0.25
(b) 0.33
(c) 0.50
(d) 0.67
(e) 2.0

Answer is B but why?

You have to show effort before getting help - forum policy.

I don't mind starting you out. If they're moving in tandem, what does that tell you? What are the forces acting on each sled?
 
the force pulling the front sled that has mass (2m) is called Applied Force
the force pulling the second sled (1m) is called Tension Force

So far I know T/F

so, F = ma
so, (T+F) = 3ma

I am lost

I don't know what to do next :(
 
gcombina said:
the force pulling the front sled that has mass (2m) is called Applied Force
the force pulling the second sled (1m) is called Tension Force

So far I know T/F

so, F = ma
so, (T+F) = 3ma

I am lost




I don't know what to do next :(

Let's take the two sleds separately.

What are the forces acting on the rear sled? You've already correctly identified the only force: tension T.

What are the forces acting on the front sled? You've only identified one of them, the applied force F. Remember that the tension of the rope also acts on this sled, but it actually retards it. So there are two forces acting on the front sled, F and -T (the minus sign signifies retardation).

Now you want to write out the equations relating the net force on each sled with its acceleration. Please answer the question I asked: what can you say about the accelerations of the two sleds if they're moving in tandem?
 
Now you want to write out the equations relating the net force on each sled with its acceleration. Please answer the question I asked: what can you say about the accelerations of the two sleds if they're moving in tandem?

1. Ok I am trying to find out the magnitude of T, how do I find that out?

T is a force , and force = mass x acceleration. The acceleration is zero and then F = (m) (0) ? i am not in the right path, i know that

2. answering to your question, if they are in tandem, i guess that means that they are at constant velocity which implies that the acceleration is ZERO
 
The sleds move on frictionless surface, and an external force acts on the system. Why should be the acceleration zero?

Imagine that the sleds are glued together so they make a single body of mass 3m. What is the acceleration of that body?

ehild
 
gcombina said:
Now you want to write out the equations relating the net force on each sled with its acceleration. Please answer the question I asked: what can you say about the accelerations of the two sleds if they're moving in tandem?

1. Ok I am trying to find out the magnitude of T, how do I find that out?

T is a force , and force = mass x acceleration. The acceleration is zero and then F = (m) (0) ? i am not in the right path, i know that
Newton's second law is $$\vec{F}_\text{net} = \sum \vec{F}_i = m\vec{a}.$$ In other words, you sum (vectorially) all the forces on the body and set that equal to the mass of the body times its acceleration.

To apply this law to a body, the first thing you have to do is identify all of the forces acting on a body. Curious3141 already told you that the only force (in the horizontal direction) acting on the rear sled is T. So Fnet=T. You now set this equal to the mass of the sled, which is just ##m##, multiplied by its acceleration, which you don't know yet. Let's call it ##a_\text{rear}##. So now we have
$$T = ma_\text{rear}.$$ Now try doing this for the front sled. First identify all of the forces on the sled. Then sum them up vectorially and set the result equal to the mass of the sled times its acceleration, which again, you don't know, so just call it ##a_\text{front}##.

2. answering to your question, if they are in tandem, i guess that means that they are at constant velocity which implies that the acceleration is ZERO
The fact they're in tandem doesn't imply they're moving at constant velocity. If the front sled is speeding up, what is the rear sled doing? How are ##a_\text{rear}## and ##a_\text{front}## related?
 
So we have REAR SLED

T = ma(rear)

and FRONT SLED

(F-T) = ma(front)
 
what is the relation of a(rear) and a(front)?

I don't know!

the front is accelerating and the one in the back is also accelerating as it is being pulled by the front sled and at the same time, it could be holding the front sled back...de-accelerating the front sled.
 
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gcombina said:
So we have REAR SLED

T = ma(rear)

and FRONT SLED

(F-T) = ma(front)
Almost right. What's the mass of the front sled? It's not ##m##.
 
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