1. Nov 3, 2009

### holezch

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

three blocks are connected on a horizontal frictionless table by a rope and pulled to the right with a force T3 = 60 N. If m3 = 30 , m2 = 20 kg and m1 = 10 kg find tensions T1 and T2.

m subscripts follow tension subscripts.. m1---t1---m2---t2----m3--- t3--->
2. Relevant equations

third law , first law.. etcera?

3. The attempt at a solution

well, I thought I could do something like 60 = forces going to the right + forces going to the left or something and then solve.. but I odn't know please help

2. Nov 3, 2009

### holezch

okay, I figured that 60 N would equal to the tensions of the other two parts plus the blocks "pulling back" I don't know how to formulate this though

3. Nov 3, 2009

### Pythagorean

step 1:

draw a free-body diagram for each mass and for the whole system

step 2:

write up the equations for the sum of the forces on each mass from the free-body diagrams (and for the whole system)

step 3:

you should have as many equations as you have unknowns (I actually didn't use the free-body diagram equation for m1).

4. Nov 3, 2009

### holezch

thank youfor the reply, am I calculating the forces on each mass so I can calculate the amount of force that is "dragging" the tension of the strings behind from becoming 60 N?

5. Nov 3, 2009

### holezch

okay , I read more about tension and apparently tension of a string is not the same at any given point. so what does it mean to find the tension of a string? isn't it different everywhere?

6. Nov 3, 2009

### Pythagorean

Well, I've assumed that the tension IS the same in this case. Since this is introductory physics, I doubt you'd have a real string with elasticity, mass, etc.

I think for this problem, you can just think of it is as an opposing force, so:

T2<---m3--->F

would lead to a free body equation:

sum of forces = (m3)a = F-T2

the only thing you don't know there is T2 and a. But you can find a by looking at the whole system as one mass being pulled with F so that:

F = Ma = (m1+m2+m3)a

7. Nov 3, 2009

### holezch

thank you for replying..60 = force of block moving in the direction of 60 N + T2 pulling block back
then continue in this manner to get all of the tensions?
but aren't we assuming that the tension of the string isn't the same everywhere, since if it were, then the string would be massless and the force 60 N would just transmit itself to the block, then the force of m3 moving would just be 60 N and the rope T2 in the other direction would be 60 N as well.. but tthis isn't the case

another thing is that, if the tensions were the same everywhere, my book says:

"Only if the acceleration a of the system is zero will we have the pairs of forces F (rope to pull), F(pull to rope) and F(block to rope), F(rope to block) the equal in magnitude"

thank you again

Last edited: Nov 3, 2009
8. Nov 3, 2009

### Pythagorean

Every problem I've done like this in an introductory course has always had tension constant. But this doesn't mean that the 60 N just transmits through. The tension in each line will be different (T1 does not equal T2) because the masses will all be moving at the same acceleration, but are different masses, so the forces connecting them must be different.

So, as I said, draw free body diagrams for each mass and the whole system of masses and get your equations. I already gave you the equation for the whole system of masses and the third mass.

9. Nov 3, 2009

### holezch

okay, I see how the strings have no mass.. thanks a lot! I solved it and everything worked out fine :)