Ah so if its the other equation, I was wrong initially
so at first it would be:
Vbc = Vbg - Vcg
meaning
Vbg = Vbc + Vcg
Then the rest is the same:
0 = MbVbg + McVcg
0 = Mb(Vbc + Vcg) +McVcg
0 = MbVbc + MbVcg +McVcg
Vcg = -MbVbc/(Mb+Mc)
= -72*55/(72+1300)
= -2.9m/s
And...
If that's true, you get
Vcg = MbVbc/(Mb+Mc)
= 72*55/(72+1300)
= 2.9
which is then negative through intuition (since its moving in the opposite direction?)
Though i suppose this works, I feel like there's a better / more accurate way to show the signs in my algebra.
Thoughts?
Since Pi = Pf,
0 = MbVbg + McVcg
I just need to express Vbg in terms of Vbc and Vcg (that is, I need to express the velocity of the ball relative to the ground in terms that I know/want to solve for):
by reference frames:
Vbc = Vbg + Vcg
so Vbg = Vbc -Vcg
Now I can sub in and solve
0 =...
Or is it simpler than this? we just assume that the made up term "Tfmax" is equal to the term "Tw" because the question states:
"The maximum forces, F, you can apply to the wrench gives a maximum torque Tw"
Thus, since Tfmax occurs when the force of friction is a maximum, it would give a...
And I know that, but I can't simply take the expression
Tf = yθ
and replace Tf with Tw when I sub in θmax: the two terms are not necessarily the same, right?
Thus, I made a term for the torque of friction when theta is at its maximum, titled "Tfmax" as a place holder
Hang on, if net work is zero, does net torque have to be zero too?
If that's true, then
0 = Tw - Tfmax
and therefore,
Tw = Tfmax , which works with my solution
Is this correct?
In that case,
Tf = y θ
so
Tfmax = yθmax ?
which rearranges to give
θmax = Tfmax/y
If I am correct, then I should integrate the first expression now:
W = (1/2)y(θmax)^2
I then sub in the equation for θmax:
W = 1/2y(Tfmax/y)^2
=(Tfmax)^2/2y
Which is very close to answer C EXCEPT...
Okay, so I gave it a try any way and set Tw = Fθ, Meaning that its antiderivative would be
1/2F(θ)^2
however, using this in my other calculations didnt get me anywhere, as it did this:
Wforce = Wfriction
1/2F(θ)^2 = (1/2)y(θ^2)
so F =y
Which hit a dead end pretty quickly, since this...
Alright, then I need to find an expression for the maximum torque's anti derivative. However, the question doesn't give me an expression for Tw in the way it does for Tf - Can you point me in the right direction please?
Ah, I apologise, I am still learning, and this problem has really got me stumped for some reason. Thanks for bearing with me.
Anyway, in that case, would the work done by the applied torque be it's anti derivative?
Okay, so if the total work is zero, shouldn't the final equation be
Wmaxforce = Wfriction
Twθ = (1/2)y(θ^2)
Tw = (1/2)yθ and at this point, I'm mostly lost, as I'm unaware of what to solve for any more, seeing that the "Wtotal" term no longer exists (its value is zero). (I might be more...