Recent content by Kermit_the_Phrog

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?

Is the momentum equation supposed to be0 = MbVbg - McVcg Because Vcg is pointed in the opposite direction to Vbg?

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 =...

Thanks again by the way - I really do appreciate the help. Enjoy the rest of your day, and you're the best.

Man, I really over thought this, huh.

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, I am with you on the integration part, but how do I relate Tw and θmax?

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...