# Force and motion problem

1. Jun 26, 2015

### J-dizzal

1. The problem statement, all variables and given/known data
In the figure here three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 33 kg, mB = 45 kg, and mC = 14 kg. When the assembly is released from rest, (a) what is the tension in the cord connecting B and C, and (b) how far does A move in the first 0.300 s (assuming it does not reach the pulley)?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c05/fig_5_E.gif
2. Relevant equations
F=ma

3. The attempt at a solution
http://s1294.photobucket.com/user/jhoversten/media/be59570a-8c6b-4612-aa51-52d1bd5d3a7d_zpsqc5pc7ce.jpg.html [Broken]

Im not sure how to solve the simultaneous equations for T or a. my attempt at solving for T was incorrect.

Last edited by a moderator: May 7, 2017
2. Jun 26, 2015

### Kinta

The first part of the question asks for the tension in the cord between blocks B and C. So, if you've run into trouble, you may want to put some of your initial assumptions under more scrutiny. Do you know for a fact that the tension in the cord is uniform throughout both segments or was that just a gut feeling?

3. Jun 26, 2015

### J-dizzal

ah crap. yea the tension in the cord is not the same through out.
I just understood the problem wrong.

4. Jun 26, 2015

### Kinta

Live and learn :)

5. Jun 26, 2015

### J-dizzal

im going to take another stab at this.

6. Jun 26, 2015

### J-dizzal

So i need to find a for the whole system because its asking for tension after the system is released right?

7. Jun 26, 2015

### Kinta

From what I've done, I am seeing that finding $a$ for the system will be beneficial, yes.

8. Jun 26, 2015

### J-dizzal

So what is the difference between the tension between box B and A and the weight of B+A?
T=ma
T=(45+14)(-9.8)=-578.2N that is the same as the weight of box B and A? So then how do i find acceleration for the system?

9. Jun 26, 2015

### Kinta

Have you written down Newton's 2nd law for the 3 masses?

10. Jun 26, 2015

### J-dizzal

here what i have so far; FA=(33kg)(a), FB=572.8N - T, FC=mg + (14kg)(-a)

11. Jun 26, 2015

### J-dizzal

FB= 572.8 + mg + TBtoC
edit: i think mg and + TBtoC are the same = -137.2N
edit again: nevermind they are different T in B=-5.4N

Last edited: Jun 26, 2015
12. Jun 26, 2015

### J-dizzal

how would I find the tension between C and B? if FC= mg + tension from B?

13. Jun 26, 2015

### J-dizzal

no

14. Jun 26, 2015

### J-dizzal

im having trouble finding the acceleration on the system because i am not getting the net forces correctly.

15. Jun 26, 2015

### J-dizzal

FA=MA a
FB=TB to A + TB to C + mg
FC=TC to B + mg
how does that look for newtons 2nd law for each box?

16. Jun 26, 2015

### Kinta

Sorry, I had to step out for a while. When writing Newton's Second Law for a thing, just put $m \vec{a} = \vec{F}^{(net)}$ where the net force would include things like gravity and tension (not the total acceleration, $\vec{a}$ which appears on the left).

Last edited: Jun 26, 2015
17. Jun 26, 2015

### Kinta

This might be right depending on your notation! Are those tensions meant to be vectors or positive scalars?

You may also find it more instructive to have $m_{A,B,C}a$ on the left side instead of $F_{A,B,C}$.
Edit: I jumped the gun a bit. Your equation for the forces on mass A isn't wrong, but go ahead and put the forces acting on it in the place of $F_A$. The previous suggestion does apply to your second two equations though.

Last edited: Jun 26, 2015
18. Jun 26, 2015

### J-dizzal

the tensions are vectors, eg. TB to A = (45kg + 14kg)a -59a

if $m_{A,B,C}a$ is on the left then what is on the right?

19. Jun 26, 2015

### Kinta

Just so that I can be sure that you're sure, can you note your coordinate system (which directions are positive and negative) and write the three equations in scalar form?

20. Jun 26, 2015

### J-dizzal

Sure, FnetA=33a, FnetB= -59a + 14a - 441, FnetC= -14a - 137.2
standard coordinate system: up positive, down negative, right positive