Finding the Tension of a cord between two masses

1. Oct 16, 2011

DanialDimitri

1. The problem statement, all variables and given/known data

Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 2.5 kg) portrayed in the figure below move on a frictionless surface and a force F = 48 N acts as shown on the 2.5-kg block.

http://www.webassign.net/sercp9/4-p-029a-mi.gif

(a) Determine the acceleration given this system.

(b) Determine the tension in the cord connecting the 2.5-kg and the 1.0-kg blocks.

and

(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block.

2. Relevant equations

ƩF = ma
F-T = ma

3. The attempt at a solution

a) 48N = ma
a = 48/5.5 <----- sum of all masses
a = 8.73 m/s^2 (which was correct).

b) I made two FBD's, one with the mass of 1kg and another with 2.5kg.
for the first mass: T=ma for the second mass: F-T = ma substitute first equation into this one: 48-(ma)= ma
48= 2(ma)
a = 24/3.5 <--- 3.5 was the sum of the two masses
a = 6.857 m/s^2
Then substitute the new value of a into the first equation: T = 3.5(6.875)
T = 24 N (This was wrong)

c) ƩF= ma
F= 1kg(8.73) <------- this acceleration was found at the beginning, and was correct
F = 8.72N (which was wrong)

Your help would be really appreciated guys.

2. Oct 17, 2011

Spinnor

Does this help?

File size:
15.9 KB
Views:
4,981
3. Feb 6, 2012

dnucuta

Great diagram! Thx a lot; I had this same question.

4. Sep 30, 2012

rbeck2

Thanks. I was making it harder than it was.