Tension in parachute cord on dragster

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    Parachute Tension
AI Thread Summary
The discussion revolves around calculating the tension in a parachute cord attached to a dragster. Participants analyze the forces acting on the dragster, including gravity, normal force, and the force from the parachute. They clarify that the tension cannot be zero, as it would imply the parachute serves no purpose. The net force is determined using the equation Fnet = ma, leading to a calculated tension of 2025 N, which matches one of the answer choices when rounded correctly. The conversation highlights the importance of understanding applied forces and their significance in physics problems.
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Homework Statement


upload_2015-11-19_20-36-54.png


Homework Equations


Fn - T = ma

The Attempt at a Solution


Fn-T=ma
810kg x 2.5m/s2 - T = 810 kg x -2.5m/s2
T=0
 
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What is this force "Fn"?
 
sorry couldn't type subscript. It is the net force.
 
angryzeena said:
sorry couldn't type subscript. It is the net force.
The net force is the sum of all the applied forces. What are the applied forces?
The acceleration is what results from the net force: ##F_{net}=\Sigma F = ma##.
 
applied forces are:
-2025 kgm/s2 from the parachute
I have a feeling we are missing other applied forces.
 
angryzeena said:
applied forces are:
-2025 kgm/s2 from the parachute
I have a feeling we are missing other applied forces.
No, that is the only force of interest. (There is gravity and the normal force from the ground, but they cancel out.)
So what is the net force? What equation does that give you?
 
ok so - tension is the force transmitted through the cord on the object. Force of gravity and normal upward force are equal so cancel out but we are trying figure out horizontal force not vertical force anyways.
We are lost...
 
angryzeena said:
ok so - tension is the force transmitted through the cord on the object. Force of gravity and normal upward force are equal so cancel out but we are trying figure out horizontal force not vertical force anyways.
We are lost...
You should ask yourself, if the tension in the parachute line is truly equal to zero, then why have the parachute attached to the dragster at all? What is the parachute doing?

angryzeena said:

Homework Equations


Fn - T = ma

Isn't the correct form of this equation Fnet = ma ?

The Attempt at a Solution


Fn-T=ma
810kg x 2.5m/s2 - T = 810 kg x -2.5m/s2
T=0

By putting the deceleration force on both sides of the equation, how else can T be anything but zero?
 
exactly what we thought - T cannot be zero as it wouldn't be doing any work.
So then we thought T = net force on the object = ma = 810 kg x 2.5 m/s2 but that seems to be incorrect? Why?
 
  • #10
angryzeena said:
exactly what we thought - T cannot be zero as it wouldn't be doing any work.
So then we thought T = net force on the object = ma = 810 kg x 2.5 m/s2 but that seems to be incorrect? Why?
How do you know it's incorrect?
 
  • #11
haruspex said that was only the force of interest not the net force. Also 2025 N is not one of the options for an answer.
 
  • #12
angryzeena said:
exactly what we thought - T cannot be zero as it wouldn't be doing any work.
So then we thought T = net force on the object = ma = 810 kg x 2.5 m/s2 but that seems to be incorrect? Why?
Does it not match any of the offered answers? The choices are only specified to two significant digits, so round your answer accordingly and then compare. What do you get?
 
  • #13
angryzeena said:
haruspex said that was only the force of interest not the net force. Also 2025 N is not one of the options for an answer.
I didn't say it was not the net force. If it's the only force of interest then it is the net force.
And 2025 N is one of the answers (see my preceding post).
 
  • #14
ah ha it is becoming clearer. I appreciate you making me think.
 
  • #15
It has been many years since physics 12 - I am helping my daughter. She and I will laugh in the morning over thinking that 2025 N does not equal 2.0x10 to the third when rounded. Many thanks...
 
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