Recent content by Ryino

Unfortunately I used, μMa meaning I probably got it wrong. I turned it in today so i don't have any of the work with me. When I get it back (Wednesday maybe?) I'll go back over and finish out the problem the right way. Hopefully I'll get partial extra credit for what I had. :smile:

Okay, i went back through it and, somehow i turned a positive into a negative in the middle of all my algebra for no reason. So now i have the same thing except Ft=2gm1m2m3(1+u)/(m2m3+m1m3+4m1m2(1+u))

Im trying to set all the masses equal and set my u to 0, and i can't find out where i messed up. I substituted like that at the point where i got my last equation and it came to 2gm/5. ... Can you see anything I'm doing wrong?from this point i would get the Ft on the left and factor/solve...

I think i got it Ft=2gm1m2m3(1+u)/(m2m3-m1m3+4m1m2(1+u))Is that correct? god i hope so Edit:: I think i can pull out an M3 from the denominator and make it m3(m2-m1)+4m1m2(1+u)

i had Ft/2m1(1+u) + Ft/2m2(1+u)=g-2Ft/m3 and i tried dividing by Ft thats how it got on the bottom With that i continued to 1/M1+1/M2=2g(1+u)/Ft - 4(1+u)/M3 I can multiply by Ft here to get it back on top on both the fractions on the left and the last fraction And originally i solved for...

I feel like I am really bad at algebra right now, so far i have 1/2m1(1+u)+1/2m2(1+u)=g/Ft-2/m3 Everything i do to this makes it feel like I am going backwards, if i multiply by ones on the right side to get a common denominator it takes that straight back to m3g-2Ft Any...

Right. that's what i have written down, forgot that two when i typed it in. Other than that do all the equations look good? I'm still working on the algebra to solve it for Ft.

Okay, for some reason i accidently substituted g for a1 and a2 on the right of the first two equations, I am reworking it now with the correct variables there, and i also see one other mistake... This time through I am using the equations M1a1=Ft-µM1a1 (1st equation) M2a2=Ft-µM2a2...

If you solve for a1 in the first equation and a2 in the second equation Plug the results into (a1+a2)=a3 and then substitute that into the the equation for M3 Using that solve for Ft And you end up getting Ft=[2m1m2m3(1+u)] / [m3(m1+m2) + 4m1m2] I think i did that right but my teacher...

Okay, so... If the Left box moves 1 meter right, and the Right box moves 1 meter left, The middle box moves down 1 meter So, (a1+a2)/2=a3

I Solved for a3 in the net force equation for M3 and got a3=9.81-(Ft1-Ft2)/M3 then I substituted the values i got for tension for the first and second box earlier and came up with a3=9.81m/s^2 - ((M1a1+µM1a1)/M3)-((M2a2+µM2a2)/M3) assuming i did my algebra right... Is that what you meant by...

If they are both treated as the same object then it would have two tensile forces on it correct? Giving you FNET=M3a3-Ft1-Ft2 and a3=g=9.81m/s^2 Both of the tensile forces will be equal I assume so, M3a=M3g-(2Ft) Right?

Homework Statement See Picture, The problem gives three boxes(M1, M2, and M3) and a few pulleys, no value for the coefficient of friction is given. None of the masses are known, and we are told to assume M1 and M3 are sliding. The objective is to find Ft (if you can't see that well...