# How can i find the acceleration for this constraint?

1. Apr 23, 2012

### yy205001

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

What is the acceleration of the 2. kg block in the figure across the frictionless table?

2. Relevant equations

F=ma

3. The attempt at a solution
g = 9.8ms^2

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2. Apr 23, 2012

### PhanthomJay

As we welcome you to these forums, you first should show more of an attempt before we can assist. You need to draw force diagrams of each block and apply newton's laws. Do you think the acceleration of the blocks are the same?

3. Apr 23, 2012

### yy205001

as long as they are in the same system, i think they have the same acceleration

4. Apr 23, 2012

### Staff: Mentor

You'll have to show more effort. What is the constraint imposed by the pulley system on the relative motions of the blocks? Did you draw FBD's for both blocks? What are the resulting force equations?

5. Apr 23, 2012

### yy205001

yea, i drew FBD for both blocks. In the x-direction, the resulting force is T. And in the y direction is (m(1kg)g-T).

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6. Apr 23, 2012

### Staff: Mentor

Okay. In your diagram you depict two "T" forces acting on the 2kg block, so why do you say the resulting force is T? How are the accelerations of the blocks related?

7. Apr 23, 2012

### yy205001

sorry, it should be 2T. Are they having the same acceleration, this is my main problem.

8. Apr 23, 2012

### Staff: Mentor

The rope and pulley system place a constraint on the motions. If the 1kg block moves down a distance d, how far will the 2kg block slide?

9. Apr 23, 2012

### yy205001

will it slides a distance d as well??

Last edited: Apr 23, 2012
10. Apr 23, 2012

### Staff: Mentor

Nope. Do a rope segment length calculation. Remember that the rope's total length is constant.

11. Apr 23, 2012

### yy205001

yea,i got d/2. Is it correct??

12. Apr 23, 2012

### Staff: Mentor

Yes, that's fine. That then will be the constraint connecting the motions of the blocks. It will apply to distance moved, velocities, and accelerations.

13. Apr 23, 2012

### yy205001

i still cannot get it

14. Apr 23, 2012

### yy205001

gneill, do you mean the acceleration of the hanging mass is twice the acceleration of the mass on the horizontal surface??

15. Apr 23, 2012

### Staff: Mentor

Yes, it must be due to the constraint...

Convert your FBD force equations to acceleration for each block. Use the constraint equation for acceleration to solve.

16. Apr 23, 2012

### yy205001

In the x-direction, the Net force is 2*T.

Let a1 be the acceleration of the mass on horizontal surface,
a2 be the acceleration of the hanging mass.
"F=m*a", so:
2*T = 2*a1
T = a1

In the y-direction, the Net force is mg-T, where m is 1kg,so:

m*g-T = m*a2
where T = a1
→m*g-a1=m*a2

Then,sub in a2=2*a1
m*g-a1=m*2*a1

So, i can find out the acceleration of a1, thus find a2?

17. Apr 24, 2012

### Staff: Mentor

Tension has units of Newtons (force). It is not an acceleration (m/s2).
Use the relationship F = ma that you've written to turn the net force into an acceleration, hence a = F/m. So that would make a1 = ?
Again you're mixing units. Forces (like m*g) are not accelerations (m/s2).

Just write the accelerations for each block separately at first; Don't try to apply the constraint until you have those expressions. You should have:

a1 = (something in terms of m1 and T)
a2 = (something in terms of m2 and T and g)

After you have them you can use the constraint equation to eliminate one of the accelerations...

18. Apr 24, 2012

### yy205001

yes, i got the correct answer finally, thank you so much!