Three blocks are connected on a table as shown in the figure below.
[/PLAIN] [Broken] <-- oopsies, apparently I don't know how to embed pictures, help please?
The table is rough and has a coefficient of kinetic friction of 0.395. The three masses are m1 = 4.21 kg, m2 = 1.14 kg, and m3 = 1.85 kg, and the pulleys are frictionless. Determine the magnitude of the acceleration of block m1, and the tension on the left cord.
I hope these are right...
F[SUB]1[/SUB] = T1 - m[SUB]1[/SUB]g = m[SUB]1[/SUB]a[SUB]1[/SUB]
F[SUB]3[/SUB] = T2 - m[SUB]3[/SUB]g = m[SUB]3[/SUB]a[SUB]3[/SUB]
F[SUB]2[/SUB] = T1 - T2 - Ff
(T1 is the left cord tension, T2 is the right)
[h2]The Attempt at a Solution[/h2]
I tried to set this up as I would if I only had 2 blocks (1 and 2).
If I only had 2 blocks:
F[SUB]1y[/SUB] = T-m[SUB]1[/SUB]g = m[SUB]1[/SUB]a[SUB]1[/SUB]y
F[SUB]2x[/SUB] = T-F[SUB]friction[/SUB] = m[SUB]2[/SUB]a[SUB]2x[/SUB] = -m[SUB]2[/SUB]a[SUB]1y[/SUB] (because a[sub]2x[/sub]=-a[sub]1y[/sub]
So I tried to account for the third block in my second equation, making it like this:
F[SUB]2x[/SUB] = T-F[SUB]friction[/SUB]-m[SUB]3[/SUB]g
I'm guessing this is where I went wrong.
I continued to use these 2 equations to eliminate T, and then I got a1 = 3.498 m/s^2, which I then used to sub into the tension equation (T-m1g=m1a1), and got 26.5N which is wrong, so I tried adding a1 and gravity and got a tension of 56N, which is also wrong!
I'm pretty sure the problem is in the equations I set up, but I don't know what would be the right equation! Any help would be greatly appreciated.
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