Average force = 1/2 maximum force?

[SaintJude]

View attachment Picture 36.png

Kap. q bank discrete

How do you know the average force = 1/2 the maximum force ? Conceptually, it doesn't make sense to me. How is the average half of the maximum ?!

Answer: 0.75 Ns

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[pfaction]

That's really weird. AUC of F*T gives you impulse.
J=F(deltaT).
J = (1/2 base height) * (1.5^-3)
J=(1/2)(1.5)(1000) * (1.5^-3)
J=500(1.5) (1.5^-3)

Wait what.

 

[SaintJude]

the base height = 1000. Check the y-axis again

And Average Force x Δ time = momentum = m Δ v

 

[MedPR]

View attachment 18966

Kap. q bank discrete

How do you know the average force = 1/2 the maximum force ? Conceptually, it doesn't make sense to me. How is the average half of the maximum ?!

Answer: 0.75 Ns

What are you talking about average force? Ft = mv. This is the definition of impulse.

The graph is F vs time where time is in milliseconds so you just convert to seconds find the area under the curve.

1000*(1.5*10^-3)/2 = 0.75Ns

 

[pfaction]

Yeah I just redid the math, I'm getting around 1+N....what the hell?

OH **** I SEE WHAT I DID WRONG, I did AUC * time! It's just the AUC.
AUC is 1/2BH. Exactly what PR did.

 

[milski]

The impulse (change of momentum) is the area under the graph of the force. Since it is a triangle in this case, the area will be 1/2 * base * height = 1/2 * Δt * Fmax. On the other hand, the average force, by definition is Favg * Δt = impulse. From here you have 1/2 Δt Fmax = Δt * Favg or Favg=1/2 Fmax. That would be true for any case where the force starts from 0, linearly increases to Fmax and then linearly decreases to 0 again.

 

[BestDoctorEver]

View attachment 18966

Kap. q bank discrete

How do you know the average force = 1/2 the maximum force ? Conceptually, it doesn't make sense to me. How is the average half of the maximum ?!

Answer: 0.75 Ns

Can you show all the solution? Edit....never mind

 

[milski]

What are you talking about average force? Ft = mv. This is the definition of impulse.

The graph is F vs time where time is in milliseconds so you just convert to seconds find the area under the curve.

1000*(1.5*10^-3)/2 = 0.75Ns

Ft = mv = P is correct only for constant F. For non-constant F you have to integrate it which is essentially what you've said - the area below the curve.

His use of Favg is correct. By definition, Favg = P/t.

 

[MedPR]

Ft = mv = P is correct only for constant F. For non-constant F you have to integrate it which is essentially what you've said - the area below the curve.

His use of Favg is correct. By definition, Favg = P/t.

Ah yea, I forgot that Favg=mv/t.

Thanks!