# Pulleys - find force to accelerate a block

1. Sep 29, 2008

### Pakbabydoll

1. The problem statement, all variables and given/known data
A force F acts to the right on a 5.65 kg block. a 2.29 kg block is stacked on top of the 5.65kg block and can slide on it with a coefficient of friction of .15 between the blocks. The table has a coefficient of friction of .18. G= 9.8 m/s^2
system is in equilibrium
Find the F required to accelerate the 5.65kg block at 2m/s^2. answer in units of N

2. Relevant equations
I am lost..... I tried but it was wrong and now I have no idea what to do.....

3. The attempt at a solution

2. Sep 29, 2008

### Staff: Mentor

Re: Pullys

Show what you've tried.

What forces act on each block?

3. Sep 29, 2008

### Pakbabydoll

Re: Pullys

So this is what I have its probably completely wrong because I did it this way and a graduate physics student did it another way but both of our answers are wrong.

N-(M1+M2)g
(2.29+ 5.65)9.8
77.812(.18)= 14.006

(2.29)(9.8)(.18)= 4.03956

F-(f1+f2)=ma
F-(14.006+4.03956)=7.94(2)

I did not even understand what the graduate student but her answer was 30.9800 (also wrong)... so do u guys have any idea? sorry it took me so long to get back but I had an orgo exam and I have been crazy studying for it.

4. Sep 29, 2008

### Staff: Mentor

Re: Pullys

This is the friction from floor. Good.

Wrong coefficient of friction.

Two problems: Wrong friction from upper block & wrong mass used. (You are applying Newton's law to the bottom block.)

5. Sep 29, 2008

### Pakbabydoll

Re: Pullys

wrong again..... sorry..... ok so I was wondering since its a pully should not we multiply the bottom acceleration by 2. umm so the bottom block would be moving twice as fast as the top block? am I making any sense? o and I have one try left to plug in the answer so can someone do this for me please? If I am wrong one more time I get negative score..:(

6. Sep 29, 2008

### Staff: Mentor

Re: Pullys

Other than the title of this thread, this is the first time you mentioned a pulley. (I assumed that the title was a mistake, since you didn't mention a pulley in the problem statement.)

Why don't you describe the problem completely, exactly as given? Got a diagram?

7. Sep 29, 2008

### Pakbabydoll

Re: Pullys

yes sir I am going to attach the hwk file. it is problem #15

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8. Sep 29, 2008

### Pakbabydoll

Re: Pullys

so any help guys?

9. Sep 30, 2008

### Staff: Mentor

Re: Pullys

Now the problem is clear. OK, so what's your attempted solution for this problem?

Redo your earlier solution, this time taking into account the string tension forces on each block and the constraint that the blocks are attached. Analyze the forces on each block and apply Newton's 2nd law.

10. Sep 30, 2008

### Pakbabydoll

Re: Pullys

That was my solution for that problem! I am stuck at that point

11. Sep 30, 2008

### Staff: Mentor

Re: Pullys

But you made no mention of tension forces or constraints. (That's why I thought your title was an error--you didn't mention a pulley in the problem statement.)

Anyway... What forces act on each block?

12. Sep 30, 2008

### Pakbabydoll

Re: Pullys

ok so there is:
Friction between top and bottom block
Friction between bottom block and the tile
Normal force but would not that just cancel with gravity?
Tension in + and - X^ directions (since all the motion is in the X^ )

Oh and I can't turn that problem in anymore it was due this morning but I really just want to get it because that was the only one from that hwk set I could not figure out. I don't know if you noticed but those problems were really hard.

13. Sep 30, 2008

### Staff: Mentor

Re: Pullys

Good. Don't forget the applied force F.

Look at each block in turn. Indentify the (horizontal) forces acting and set up Newton's 2nd law. You'll get two equations (one for each block). If you combine them you can solve for F.

How are the tensions acting on the blocks related? The accelerations of the blocks?

Hard problems are good for you! :tongue:

14. Nov 7, 2008

### physics_liker

Re: Pullys

g=9.8 ms^-2
a=2 ms^-2
u1= coefficient of friction between the two blocks
The upper pulley acc. at 1/2 of the acc. of the bottom block.

So I got 2T-2.29*g*u1=2.29*a/2
T=2.83 N

F-5.65*g*u2-T=(5.65+2.29)*a
F=20.73 N

15. Nov 10, 2008

### Staff: Mentor

Re: Pullys

Please do not post complete solutions to problems.
OK.
This equation is incorrect.

16. Nov 11, 2008

### murat

Re: Pullys

fırstly
2T-2.29*9.8*0.15=(a=1*2.29) T=2.82815

F-(2.29*0.15*9.8)-T-((5.65+2.29)*9.8*0.18)=(a=2*5.65)

F=30.61450600

İS MY ANSWER RIGHT?