Finding Acceleration & Tension in a Frictionless Pulley System | Homework Help

[nns91]

Homework Statement

A bos of mass m2= 3.5 kg rests on a frictionless horizontal shelf is attached by strings to boxes of masses m1= 1.5 kg and m3= 2.5 kg, which hang freely. Both pulleys are frictionless and massless. The system is initially held at rest. After it is released, find (a) the acceleration of each of the boxes, and (b) the tension in each string.

If you are confused about the graph, imagine the the m2 is in the middle with 2 strings on 2 sides, at 2 ends of the shelf there are 2 pulleys and m1 hangs on the left of the shelf and m3 hangs on the right of the shelf.

Homework Equations

F=ma

The Attempt at a Solution

I tried to get the Tension at rest but it not going to do anything with the final answer.

 

[nns91]

Guys, I need some help really quick

 

[aimslin22]

Ok, this is rather long.

First you need to find the acceleration of the masses. The acceleration of all the masses is the same. So, for now look at the three masses as one big box (added them up to together). Then draw the forces pulling them each way. (There should be two).

Then use F= ma

subtract the smaller force from the larger force, because the force is net force, and solve:

so it should be set up like this:

winner force- loser force = mass(of entire big blog) * a

 

[PhanthomJay]

Guys, I need some help really quick

You are going to have to show some work. You will need to identify the forces acting on each block, and apply Newton's 2nd law to each. Are you familiar with free body diagrams?

 

[einstein_a_go_go]

hi,

Draw the diagram

draw a free body diagram for each box (remember the tensions in each string may be different).

If the strings are taut and unstretchable, what can you say about the accelerations of each body?Now you should have lots of equations! Solve!

 

[nns91]

How do I find the acceleration ? Does it involve tension here ??

 

[nns91]

I did draw FBD for each box and creates some 2nd law equation:

T1-m1g=m1a1
m3g-T3=m2a2
T3-T1=m2a2

What do you guys thinks ?

 

[aimslin22]

I posted how to find the acceleration. forget about the tension until you find the acceleration. (you need acceleration to find the tension)

 

[nns91]

I mean how do you find the force since we don't know a ?

 

[aimslin22]

you can't

 

[nns91]

so how can I use your formula:
winner force- loser force = mass(of entire big blog) * a

to find a ?

 

[aimslin22]

find a, then find tension. It's the only way.

 

[nns91]

I know that so how do you find a ? Which formula do you use ?

 

[aimslin22]

winner force- loser force = mass(of entire big blog) * a

look at my first post

 

[nns91]

That's what I thought you wrote. So I can find the mass. How about the winner force and loser force ??

 

[aimslin22]

the winner force and the loser force come from the boxs hanging down. Since they are hanging, they have weight, so: mg

 

[nns91]

so we ignore the tension there ?

Now I get it. Sorry I am a little too dumb tonight.

 

[aimslin22]

it's alrgiht. Tell me when you get the acceleration, and then I'll help with the tension. (there are two)

 

[nns91]

I got 1.31 m/s^2. Is that right ?

Now what's next ?

Do we use 2nd law again as:

T1-m1g= m1a ?

m3g-T3=m3a ?

 

[aimslin22]

sorry, I actually have not done the calculations. but the set up for the next part looks right!

 

[PhanthomJay]

I did draw FBD for each box and creates some 2nd law equation:

T1-m1g=m1a1
m3g-T3=m2a2you meant to say m3g - T3 = m3a3[/color]
T3-T1=m2a2

What do you guys thinks ?

Except for the typo in your second equation, which you have corrected in your last post, your equations are good. And since the accelerations of each block are equal, a1=a2=a3, so just call it a. Solve the three equations with the 3 unknowns (T1, T3, and a) and you're done. Drawing good free body diagrams are important in solving most problems involving Newton's laws.

 

[nns91]

Thank you. That helps a lot. I was kinda dummy when not thinking about that the acceleration is same for 3 of them.

What if in another problem, there is a m1 on an inclined plane (with angle x), on top of the inclined plan, there is a pulley and m1 is connected to m2 (hanging freely) through a string through the pulley.

How do I solve this one ?

I know that acceleration for m1= g sin x. What will be the acceleration of m2 ? Will it be the same ??

 

[PhanthomJay]

Thank you. That helps a lot. I was kinda dummy when not thinking about that the acceleration is same for 3 of them.

What if in another problem, there is a m1 on an inclined plane (with angle x), on top of the inclined plan, there is a pulley and m1 is connected to m2 (hanging freely) through a string through the pulley.

How do I solve this one ?

I know that acceleration for m1= g sin x. What will be the acceleration of m2 ? Will it be the same ??

As long as the blocks are connected by a taut inextensible string wrapped around a massless frictionless pulley, the magnitudes of their accelerations will be the same. However, the acceleration for m1 is not gsinx; you have to consider the string tension as well in your free body diagram for each block. And depending on the values of m1 and m2 ,m1 might be moving down the plane , or up the plane, or not moving at all!

 

[nns91]

If so, how can you find ound the acceleration for m1 ?

 

[PhanthomJay]

If so, how can you find ound the acceleration for m1 ?

Just like you did in the previous problem, draw a FBD of each block. You willend up with 2 equations with 2 unknowns (T and a). Solve for each. If the block are not moving, 'a' will come out to zero. Otherwise, if 'a' comes out to a positive or negative number, it's moving one way or the other.

 

[nns91]

Oh yeah, I know what you mean.