# Balloon in a Truck

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#1
A balloon is held by a light string in the middle of the back of a very powerful truck. The back of the truck is sufficiently large that the balloon cannot hit the sides if it sways forwards or backwards on the string. The back of the truck is also completely enclosed and devoid of any air (it's a vacuum). The truck now accelerates rapidly forwards. What is the motion of the balloon relative to the truck when the truck is accelerating? Does the balloon sway forwards, backwards or remain at the same position?

It intuitively seems that the balloon will sway backward as the bottom of the string moves forwards faster than the balloon itself. But I find it hard to put it forward formally.
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2 years ago
#2
(Original post by esrever)
A balloon is held by a light string in the middle of the back of a very powerful truck. The back of the truck is sufficiently large that the balloon cannot hit the sides if it sways forwards or backwards on the string. The back of the truck is also completely enclosed and devoid of any air (it's a vacuum). The truck now accelerates rapidly forwards. What is the motion of the balloon relative to the truck when the truck is accelerating? Does the balloon sway forwards, backwards or remain at the same position?

It intuitively seems that the balloon will sway backward as the bottom of the string moves forwards faster than the balloon itself. But I find it hard to put it forward formally.
When at rest, draw a free body diagram of the balloon. It will have weight, mg, acting on it downwards, and tension from the string.
When the truck accelerates with respect to the balloon, we can think of it two ways and here's where it gets fun
1) we can consider the additional acceleration as an additional gravitational field. In other words the resultant weight of the balloon is no longer downwards, but becomes the vector sum of the acceleration due to gravity and acceleration with respect to the truck. That's because in the truck's point of view (I hesitate to say frame of reference because it's accelerating hence you could determine which object truly has a force acting on it, but in this simplified situation there is little difference) the balloon has a resultant acceleration acting backwards on it equal and opposite to the truck's.
Ok how draw this? It's exactly the same as if you were to tilt the truck in the original gravitational field. Hence the balloon's string will tilt back and make an angle to the vertical which you can work out given the truck's acceleration and g.
2) a simpler argument is inertia. The string is tied to the truck but the balloon is not actually accelerating alongside the truck, since it is in a vacuum. The balloon is its own system so will refrain from moving at first, until the tension in the string, which by this point is angled, pulls it forwards. (The forwards force on the balloon is Tsin(theta) where T is the tension and theta the angle to the vertical).

Hope this helps. You can try Googling frames of references too; this example comes up pretty often.
Last edited by DeBrevitateVitae; 2 years ago
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#3
(Original post by DeBrevitateVitae)
This is a very interesting example of frames of references.
When at rest, draw a free body diagram of the balloon. It will have weight, mg, acting on it downwards, and tension from the string.
When the truck accelerates with respect to the balloon, we can think of it two ways and here's where it gets fun
1) we can consider the additional acceleration as an additional gravitational field. In other words the resultant weight of the balloon is no longer downwards, but becomes the vector sum of the acceleration due to gravity and acceleration with respect to the truck. Ok so how to draw this? It's exactly the same as if you were to tilt the truck in the original gravitational field. Hence the balloon's string will tilt back and make an angle to the vertical which you can work out given the truck's acceleration and g.
2) a simpler argument is inertia. The string is tied to the truck but the balloon is not actually accelerating alongside the truck, since it is in a vacuum. The balloon is its own system so will refrain from moving at first, until the tension in the string, which by this point is angled, pulls it forwards. (The forwards force on the balloon is Tsin(theta) where T is the tension and theta the angle to the vertical).

Hope this helps. You can try Googling frames of references too; this example comes up pretty often.
The first idea is really cool .

I just want to clarify the second idea. So the balloon remain at rest initially because it doesn't have an external force acting on it. As a result, it moves back with reference to the truck until the string prevents it from moving back any further. Is this correct?
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2 years ago
#4
(Original post by esrever)
The first idea is really cool .

I just want to clarify the second idea. So the balloon remain at rest initially because it doesn't have an external force acting on it. As a result, it moves back with reference to the truck until the string prevents it from moving back any further. Is this correct?
I made a small error explaining that; you can't assume the force suddenly starts acting. I think it's correct to assume no external force on the balloon at the beginning. However past t=0, as the balloon moves back, the component of tension in the horizontal direction increases as the angle increases. The angle will stop increasing once this horizontal component provides an acceleration equal to the truck's. A graph of horizontal force against time should look a bit like a sine curve between 0 and theta.
Last edited by DeBrevitateVitae; 2 years ago
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#5
(Original post by DeBrevitateVitae)
I made a small error explaining that; you can't assume the force suddenly starts acting. I think it's correct to assume no external force on the balloon at the beginning. However past t=0, as the balloon moves back, the component of tension in the horizontal direction increases as the angle increases. The angle will stop increasing once this horizontal component provides an acceleration equal to the truck's. A graph of force against time should look a bit like a sine curve between 0 and theta.
Makes sense. Thanks .
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2 years ago
#6
(Original post by DeBrevitateVitae)
I made a small error explaining that; you can't assume the force suddenly starts acting. I think it's correct to assume no external force on the balloon at the beginning. However past t=0, as the balloon moves back, the component of tension in the horizontal direction increases as the angle increases. The angle will stop increasing once this horizontal component provides an acceleration equal to the truck's. A graph of horizontal force against time should look a bit like a sine curve between 0 and theta.
I would say there is no net force or no resultant force instead of no external force.
If the balloon is hanging, there is tension in the string (string exerts a force on the balloon) and there is a gravitational force acting on the balloon.
If the balloon is resting on the floor of the back of the truck, then there is a normal force and gravitational force exerting on the balloon. There may be friction.

I would add the following to your analysis in post # 2.
Newton's second law needs to be modified if it is to be used in an accelerated frame of reference.
The observer in the accelerated truck frame of reference would introduce a force (which we know to be fictitious) in the horizontal direction to balance the horizontal component of tension and claims that the net force on the balloon is zero.

In using inertia, the frame of reference is the inertial frame of reference.
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