# Forces help -- 2 masses, a pulley and friction...

1. Jul 1, 2017

### deuce123

1. The problem statement, all variables and given/known data
Determine a formula for the acceleration of the system shown in the figure(Figure 1) in terms of mA, mB, and the mass of the cord, mC. Assume f to be the fraction of the cord that is hanging down between mB and the pulley, and μk to be the coefficient of kinetic friction between mass mA and the table.
Express your answer in terms of the variables mA, mB, mC, μk, f, and appropriate constants.

2. Relevant equations
F=ma

3. The attempt at a solution
I've been trying to solve this for the past hour but I can't seem too get it. I know that for mass B---(mB+f)g-Tb=(mA+mB+mC)a-- where Tb is the tension in the cord. Also for A----Ta-μk(mA*g)=mAa------ & n=mA*g--------From here on I've been stumped. Whatever approach I took to solve the problem didn't end up working. Can somebody please push me in the right direction. Also, I'm unsure if my force equations are correct, somebody please help.

2. Jul 1, 2017

### BvU

What is the difference between TA and TB ?

3. Jul 1, 2017

### deuce123

I wanted to represent weight, would it rather be f/mC+mB? And b/c they stated we have mC, the tension in the rope varies from place to place I assume. I'm just going based off what it think is right, this is the first problem I've done with a rope with a mass.

4. Jul 1, 2017

### BvU

No. The two terms must have the same dimension. f/mB has [mass]-1

Good point (I must admit I overlooked it ) .
For a massless rope the tension is the same all along the rope. Here you can draw a free body diagram for the fraction f of the rope from pulley to mB. For the fraction 1-f between mA and the pulley you do the same.
The rope does not stretch, so there is only one acceleration.

5. Jul 1, 2017

### deuce123

How would the forces on the ropes differ from the boxes? I'm really confused about how the forces work out and how the lengths contribute into the equations

6. Jul 1, 2017

### BvU

Gravity works on the mass of the rope from pulley to mB.

Did you draw/sketch the free body diagrams ? They help you set up the equations.

7. Jul 1, 2017

### haruspex

It might be easier to think of the rope as massless, and divide it into four lengths. First length is from mass A to a mass representing the mass of the horizontal portion of rope. Next length is from there to the pulley, etc.

8. Jul 1, 2017

### Staff: Mentor

Please draw three separate free body diagrams: mass A, mass B, and the rope. Show the forces acting on each of these free bodies. What is your force balance on mass B now?

9. Jul 2, 2017

### deuce123

I'm 100% sure I've got the FBD for mass A down, for the FBD of the rope, I'm assuming only the portion from A to the pulley is the significant portion, I got T-friction=(mA+f/L(mC))a where T is the tension made from mass B and the rope from that side and f is the fraction of the rope from A to the pulley, and L is the length of the rope. I'm not sure if this is correct, and also for B I got [f/L(mC)+mB]g-T=(mA+mB+mC)a where this f is the fraction from the pulley to B and this tension is the tension from the pulley to B. I'm pretty confident I've done something wrong haha, please push me in the right direction

10. Jul 2, 2017

### haruspex

Why? The vertical portion has weight, contributing to the motive force, and mass, contributing to the inertia.
The question defines the hanging fraction as f.
f is a fraction already, not a length. Besides, L is not a given variable, so cannot feature in the answer.

11. Jul 2, 2017

### deuce123

I tried to contribute the vertical portion to the FBD for massB, but I'm not sure how to divide up each length in the rope with the variables given. I've almost given up on this problem

12. Jul 2, 2017

### haruspex

I later noticed you do include f in the equation for the vertical motion, so I guess you only meant that only the horizontal part is relevant to the horizontal motion.
On the left, (apart from the f/L which should be just f) you have the net force acting on B and the vertical portion of rope. But on the right you have included other masses. You need to be consistent with your selection of subsystem when writing ΣF=ma equations.

13. Jul 2, 2017

### deuce123

OH, I've been thinking about it wrong for the masses aswell( for the ma part) I thought that all the mass contributed to that portion, but it doesn't. Thank you I'll give it a shot!

14. Jul 2, 2017

### BvU

Do you have a way to post the free body diagrams ? I'm not reassured by
since that doesn't involve the rope, except for the tension the rope exerts on mB

15. Jul 2, 2017

### BvU

Ponder:
What is it in this excercise that needs to be accelerated ? (did you already conclude that there is only 1 acceleration ?)
What is the driving force for this acceleration ?

16. Jul 2, 2017

### deuce123

for A I have n=Fg, T1-friction=mAa
For B I have (fmC+mB)g-T2=(fmC+mB)a The driving force for the acceleration is the weight of B+ the weight of the rope from the pulley to B. Also isn't acceleration the same for all parts connected?

17. Jul 2, 2017

### BvU

What is n ? Normal force ? (we are used to capital N). And you know Fg (if that is the force from gravity) So you can already write 'friction' in terms of know variables (*)
This is not for B. If T2 is the tension at B, then you simply write mBg - T2 = mB a
Yes !
Yes !

It's good to work in terms of (preferably known) variables a long as you can:
It helps you check dimensions in every step
Often several of them cancel so you have less to calculate --> less errors and better accuracy
Don't use more different variables than necessary

Use subscripts: an expression with mB is easier to read than one with mB
(ultimately using $\LaTeX$ is even better...)​

18. Jul 2, 2017

### haruspex

It would help if you were to make clear exactly where on the ropes these tensions refer to.
I agree with those equations if T1 is the tension directly adjacent to mA but T2 is the tension at the top of the vertical portion of rope.
Now you need an equation relating the two.

19. Jul 2, 2017

### Staff: Mentor

My force balance on A agrees with yours: $T_A-\mu_k m_Ag=m_Aa$
My force balance on B does not agree with yours: $m_Bg-T_B=m_Ba$
Now, what is your force balance on the cord?

20. Jul 2, 2017

### deuce123

For the cord connecting A to the pulley; (mB+fmC)g-μmAg=(1-f)mCa------- is this correct? the force of gravity of the cord thats connected to be and the mass B itself push the acceleration, and the friction from A holds opposed the acceleration.