B I just found a new way to calculate Force

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A new formula for calculating force, F = (m/t)v, is proposed as a more efficient method that does not require acceleration. However, this approach leads to inconsistencies with established physics, particularly regarding how force behaves over time. Critics point out that the formula fails to align with experimental observations, such as the fact that force increases with time during free fall. The discussion highlights the importance of ensuring that any new formula is consistent with physical laws and experimental data. Ultimately, while the proposed formula offers an alternative perspective, it does not accurately reflect the realities of force dynamics.
John Clement Husain
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I found a new way to calculate force without getting the acceleration when you have velocity, mass, & time.
F = (m/t)v
I just find this more efficient than getting the acceleration, why not use one formula.
 
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But as acceleration is change of velocity per time, you get ##F=(m/t)v=(mv/t)=m(v/t)=ma##, which is acceleration again.
 
So according to your formula, force always decreases as time increases. Is this what you observe?
 
Dale said:
So according to your formula, force always decreases as time increases. Is this what you observe?
It seems so, yes
 
fresh_42 said:
But as acceleration is change of velocity per time, you get ##F=(m/t)v=(mv/t)=m(v/t)=ma##, which is acceleration again.
yeah, I just stated that I don't have to go to a=v/t anymore
 
John Clement Husain said:
It seems so, yes
No, it doesn't. In fact, as an object falls from a great distance the force increases as time increases, in direct opposition to your formula.

When you write a formula in physics, it is important to check if it is consistent with experiment. Yours is not.
 
Dale said:
No, it doesn't. In fact, as an object falls from a great distance the force increases as time increases, in direct opposition to your formula.

When you write a formula in physics, it is important to check if it is consistent with experiment. Yours is not.
oh, I see...
also in my formula, force increases due to it's velocity. The faster the object, the greater the force?
but the formula does not contend with reality though, I shall take a note on that, thanks
 
John Clement Husain said:
oh, I see...
also in my formula, force increases due to it's velocity. The faster the object, the greater the force?
but the formula does not contend with reality though, I shall take a note on that, thanks
Also your understanding of force only works for constant accelerations.

constant a and constant m:
The force equation is F=ma, we could know the change in velocity Δv and the amount of time it took for that change to occur Δt. Then acceleration is a constant a=Δv/Δt. You could them formulate an equation F=(m/Δt)*Δv but this is trivial.

variable a and m:
Force eq. is F=dp/dt. this is the rate of change of momentum using differential calculus. You could have a changing m in a rocket, where the acceleration and velocity change as the rocket flies up, but the mass decreases as huge containers of fuel are emptied. Since p=mv, the force equation could read F=d(mv)/dt and your discovery is actually quite meaningful as you have noticed that Forces aren't limited to changes in velocity over time only.
 
John Clement Husain said:
also in my formula, force increases due to it's velocity. The faster the object, the greater the force?
Yes, good observation. It would mean that objects at rest would get stuck because v=0 so F=0. You could throw a ball up and have it never come down but just rest at the apex forever, and you would never be able to get out of bed in the morning.
 
  • #10
John Clement Husain said:
I found a new way to calculate force without getting the acceleration when you have velocity, mass, & time.
F = (m/t)v
I just find this more efficient than getting the acceleration, why not use one formula.

You could have written:

##F = m\frac{dv}{dt}##
 
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