What is the acceleration of the masses in this physics homework problem?

[jbgibson]

I think someone may have previously asked for help on a problem similar to this one:

There are three blocks connected by a string. m1 is 0.5kg, m2 is 1.7kg.
m1, and m2 are connected at the top of a surface and m3=2.9kg, is hanging of the end of a pulley. looks kinda like this:

[ m1 ]----[ m2 ]-----O
xxxxxxxxxxxxxxxxxxxxx|
xxxxxxxxxxxxxxxxxxxxx|
xxxxxxxxxxxxxxxxxxx [m3]

a) find the acceleration of the masses shown in the figure.

I realize that the only force acting on m1: F=T= m1a, and I think the force acting on m3: m3g-T=m3a. I am cofused about what to do with m2. Any help is greatly appreciated.

JB Gibson

 

[Doc Al]

Realize that each piece of string--the one between m1 & m2 and the one between m2 & m3--will have a different tension.

 

[jbgibson]

Realize that each piece of string--the one between m1 & m2 and the one between m2 & m3--will have a different tension.

Would I be correct in my assumptions:
F12=T12=a(m1+m2)
F23=m3g-T23=m3a

m3g = a
m1+m2+m3

 

[Doc Al]

I don't understand what you've done. Try this approach:
(1) Identify the forces on each mass
(2) Write Newton's 2nd law for each mass

You'll get three equations that you can solve together. Hint: Assume that the acceleration is "a" to the right and down.

 

[jbgibson]

I don't understand what you've done. Try this approach:
(1) Identify the forces on each mass
(2) Write Newton's 2nd law for each mass

You'll get three equations that you can solve together. Hint: Assume that the acceleration is "a" to the right and down.

That's where my confusion begins! How do I derive these equations? I thought I was ok on the first and last equations, but evidently not. The first and last equations are as follows:

F1=T=m1a
F3=m3g-T=m3a

Thaks for all the help!

 

[Doc Al]

F1=T=m1a
F3=m3g-T=m3a

As I mentioned earlier, the tensions on the strings are different. I'd write these two equations like this:
[tex]T_1 = m_1 a[/tex]

[tex]m_3 g - T_2 = m_3 a[/tex]

Now write the equation for the second mass.

 

[jbgibson]

As I mentioned earlier, the tensions on the strings are different. I'd write these two equations like this:
[tex]T_1 = m_1 a[/tex]

[tex]m_3 g - T_2 = m_3 a[/tex]

Now write the equation for the second mass.

I'm having a hard time deriving the equation for the second mass. I'm not sure what to do. I know that the gravitational force and the normal force cancels out, and you're left with the tension force and the friction force. How do you apply this information?

 

[Doc Al]

Unless the problem states otherwise, assume the surfaces are frictionless. (If friction is involved, you'll need to include friction in the equations for both m1 and m2.)

So all you need to worry about are the two tension forces that act on m2. Give it a shot.

 

[jbgibson]

Unless the problem states otherwise, assume the surfaces are frictionless. (If friction is involved, you'll need to include friction in the equations for both m1 and m2.)
So all you need to worry about are the two tension forces that act on m2. Give it a shot.

Let me get this straight! Here are my equations:

T1=m1a
T2=m2a-m1a
m3g-T2=m3a

Just a guess for clarification. Thanks again!

 

[Doc Al]

Let me get this straight! Here are my equations:
T1=m1a
T2=m2a-m1a
m3g-T2=m3a

You still haven't got the correct equation for m2. Answer these questions:
(1) What horizontal forces act on m2?
(2) What's the net force on m2?
Now apply Newton's 2nd law ([itex]F_{net} = m a[/itex]) to mass 2.

 

[jbgibson]

To make things a little clearer, I guess I need to know what is the equation for m2. I not asking anyone to do the problem for me or derive any solutions. I just don't see the 2nd equation!

 

[jbgibson]

You still haven't got the correct equation for m2. Answer these questions:
(1) What horizontal forces act on m2?
(2) What's the net force on m2?
Now apply Newton's 2nd law ([itex]F_{net} = m a[/itex]) to mass 2.

I think I get it! Correct me if I'm wrong:

T1=m1a
T2-T1=m2a
m3g-T2=m2a

This would leave me with a final equation of:

m3g=a(m1+m2+m3)

and now I would solve for "a"

Is this correct?

 

[Doc Al]

Now you're cooking!

(You did have a typo in your third equation.)

 

[jbgibson]

Thanks Doc for all of the help and letting me do the problem. I'm grateful for the help!