Relating terminal velocity, force of friction.

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Discussion Overview

The discussion revolves around the relationship between terminal velocity and the forces acting on an object, specifically the force of gravity and the force of friction. Participants explore the reasoning behind how these forces balance at terminal velocity, with a focus on a specific problem involving a block sliding down an incline.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how the force of gravity can equal the force of friction at terminal velocity, seeking a conceptual understanding rather than just mathematical equations.
  • Another participant clarifies that at terminal velocity, forces balance, implying that friction must counteract gravity, leading to a state of no acceleration.
  • A third participant presents the equation for net force, indicating that at terminal velocity, the net force is zero, which supports the idea that friction opposes gravity.
  • One participant challenges the premise of the question, arguing that sliding friction is not typically dependent on speed, suggesting that the concept of terminal velocity may not apply in the same way for sliding friction.
  • A later reply expresses gratitude for the clarification regarding acceleration being zero at terminal velocity, indicating a shift in understanding.

Areas of Agreement / Disagreement

Participants generally agree that at terminal velocity, the forces balance, but there is disagreement regarding the nature of friction and its dependence on speed, particularly in the context of the problem presented.

Contextual Notes

Some participants note that the question may not accurately reflect physical principles, particularly regarding the speed dependence of sliding friction, which remains unresolved.

bravoghost
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I was working on problems for the MCAT and came across this below. After a few minutes working, I had couldn't come up with the solution. Usually the answer clears up any questions I have, but I'm having trouble figuring out how the force due to gravity matches the force of friction. I thought that the force of friction was something to subtract from the total force available for an object. I don't understand how the force due to gravity equals the force due to friction. How do those two relate? (I understand the mathematics behind it, I just don't understand the reasoning, so please don't just show a bunch of equations and conclude "Here: the math shows it."

Question:
A block slides down a surface at an angle of 30° to the horizontal at its terminal velocity of 5 m/s. If the block masses 12 kg, what is the coefficient of kinetic friction of the surface?

Answer: (B) =1/√3

Explanation: As the block travels down the incline at its terminal velocity, the force due to gravity matches the force of friction (hence terminal velocity). The equation, therefore, can be set up as μN = mg(sin θ) where N is the normal force and equal to mg(cos θ). Thus our equation is μmg(cos θ) = mg(sin θ). mg cancels on both sides leaving us with (sin θ)/(cos θ) = μ. Given the angle of 30°, we can calculate that μ = 1/√3 making (B) the correct answer.
 
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When something is moving at terminal velocity, then it is moving at a constant speed, which means that the forces balance so the friction has to balance gravity does't it? Either that or there's another force.

In situations such as air resistance, the friction depends on speed ... so it starts out too small to prevent acceleration so [itex]mg-\mu(v)N=ma[/itex] and increases until it is the same as gravity ... in which case you write [itex]mg-\mu(v_t) N = 0 \Rightarrow mg=\mu(v_t)N[/itex].
 
Last edited:
Fnet=ma
a=0 for teminal velocity.

Fg+(-Ffr)=0

Ffr is negative since it opposes the direction of the motion.
 
Clearly whoever made up the question wasn't a physicist though. Sliding friction is not typically dependent on movement speed, so the whole concept of terminal velocity is sort of silly. Someone could start that block moving at, say, 8 m/s and it would stay at that speed as well.
 
Simon Bridge! THANK YOU. That makes so much more sense knowing that the terminal velocity means that the acceleration has dropped to 0. It was that one little piece that made the difference. I got it now.
 

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