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Tension and Acceleration on Blocks

  • #1

Homework Statement


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.

[h2]Homework Equations[/h2]
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.
 
Last edited by a moderator:

Answers and Replies

  • #2
1,137
0

Homework Equations


I hope these are right...
F1 = T1 - m1g = m1a1
F3 = T2 - m3g = m3a3
F2 = T1 - T2 - Ff
(T1 is the left cord tension, T2 is the right)
You would know that as m1>m3 so m1 will go down ...
therefore make changes in your eqn:
F1 = T1 - m1g = m1a1 accordingly
 
  • #3
F1 = T1 - m1g = -m1a1

Thanks! I tried the problem again and I'm still wrong, but less wrong! haha. Can you help me out with the second equation? I think that's where I'm off.
 
  • #4
1,137
0
No ...

ok ... suppose that a ball is acted by forces 500N in south and 490N in north ... how do you write eqn for finding the acceleration?
 
  • #5
F=-500+490=-10=ma
a= -10/m ?
 
  • #6
1,137
0
Ok...

A little tip, in ques like this you need not take one direction as positive... you can also take direction along lotion pf objects as positive

And look at the figure ... do you see any relation between a1 and a3??
 
  • #7
Thanks! I got it now!
I found the net force on block 2 (m1g-m3g-friction on the second block) and then divided it by all the masses.
thanks for your help!!
 

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