# Two Blocks with friction

1. Oct 28, 2004

### Mivz18

Ok, I have this problem I've been working on forever, but can't figure out the last part of the question.

A block of mass m1 = 2 kg and a block of mass m2 = 3 kg are tied together and are pulled from rest across the floor by a force of P = 30 N. The coefficient of friction of the blocks with the floor is µ = 0.1.

First it asked what the acceleration of the two blocks were, and I figured that out to be 5.02 m/s ^2 . Then the second part asks what is the tension of the string between the two blocks. This is where my trouble begins.

I know that [/SIZE]Fnet = T on m1, unless I'm mistakenly not counting the friction. However, with the force of 30 N to the right and 4.9 N of friction to the left in the system, that is not accounted for in the Fnet of 10.04 N for m1 and 15.06 for m2 that I obtained. If T is the only force on m1, wouldn't T = 10.04 N ?? Please help!
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2. Oct 28, 2004

### NateTG

You could draw an FBD for both of the blocks with seperate tension values for both of the sections of string. Then you should be able to set up, and solve, a system of equations.

3. Oct 28, 2004

### Sirus

You know the acceleration and mass of each block. You also know the applied force and coeff. of friction. Since $F_{net}=ma$, can you develop the equation to include $F_{tension}$, which you can then isolate to answer the question?

4. Oct 28, 2004

### Mivz18

I had tried developing an equation but it doesn't seem to work like I had stated. Since in m1 Fnet = T, then T should = 10.04 N ? The online program I am using doesn't agree with that answer, so I'm kind of stumped here. I have drawn FBD for both blocks but haven't been able to figure it out. HELP!?!?

5. Oct 28, 2004

### Staff: Mentor

Start by indentifying all the horizontal forces on m1. (The tension in the string is not the only force.) Apply Newton's 2nd law.

6. Oct 28, 2004

### Mivz18

Ok, maybe I'm missing something but this is what I get:

on m1, of 2 kg, you obtain two forces, that of friction to the left, and that of tension to the right. Therefore, Fnet = T - friction, or T- friction = ma . From the acceleration I obtained in the first part, 5.02 , T - friction = 10.04.

on m2, of 3 kg, I obtain three forces, that of the applied 30 N to the right and that of friction and tension to the left. Fnet = 30 -T - friction , or 30 - T -friction = ma . From the acceleration of 5.02 i get that T + friction = 14.94 .

Then I add the two equations to solve for the variable, Tension, that I'm looking for.

T - friction = 10.04
T + friction = 14.94

2T = 24.98
T = 12.49

However, the online program disagrees with that answer. Did I calculate something wrong, or am I assuming that the two T's are equal and they really aren't?

7. Oct 28, 2004

### Staff: Mentor

Right, but "friction" is not an unknown! Figure it out. Then solve for T. And you're done.

Right, but not needed. Once again "friction" can be figured out.

Big mistake! The two things you label "friction" are not the same!

8. Oct 28, 2004

### Mivz18

nevermind, lol, I found my mistake. I was in the wrong in adding the equations. Instead,

T - friction = 10.04
T = 10.04 + friction = 10.04 + 1.96 = 11.98

T + friction = 14.94
T = 14.94 - friction = 14.94 - 2.94 = 12

So basically, both are about 12 N , and the online program accepted it. Thanks everyone for your guidance.

9. Oct 28, 2004

### Staff: Mentor

tension is the same throughout the string

Just to be clear: There is only one tension in the string. No need to solve for it twice! (Any difference in the two answers is just due to rounding errors.)

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