# Homework Help: Dynamics of a particle

1. Mar 25, 2012

### HWGXX7

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

2. Relevant equations

I converted the given data to SI units, because of my native calculation system.
I used an index for particles and named rope force BC: $$T_{BC}$$

3. The attempt at a solution

Kinematic expression for particle B:
$$a=constant=1,22 \frac{m}{s^{2}}$$
$$v=a.t$$
$$x(2)=a.\frac{t^{2}}{2}=2,44 m$$

Dynamic expression for particle B:
$$m_{B}.a_{B}=44,5+P-T_{BC} => 4,54 kg.1,22\frac{m}{s^{2}}= 44,5+P-T_{BC}$$

Finding unknown rope force BC form dynamic expression of particle C:
Acceleration of particle C is the same as particle B
$$-5,54=44,5-T_{BC}$$
$$T_{BC}=-50,.04N$$

Apply this in equation for particle B gives: $$5,54=44,5+P-50,04 => P=11,08N$$

(Correct) solution has to be: 7,37 N

What did I do wrong?

2. Mar 25, 2012

### Staff: Mentor

Notice that your solution doesn't include any effects from body A. That should make you suspicious

If pulley D accelerates downwards (as body A accelerates upwards), then this will affect the tension you find. If the pulley is accelerating downward then the whole "subsystem" consisting of the pulley with body B and body C will share the same net acceleration. One effect of this will be to reduce the effective gravitational acceleration acting on B and C. If we call the downward acceleration of the pulley A1, then the effective gravitational acceleration operating on B and C will be g - A1.

3. Mar 28, 2012

### HWGXX7

I reviewed the problem and came to follow set of equations:

Subsystem with particle B and C (and the pully) has an acceleration $$a_{1}$$
Newton second law:
$$+9,08.a_{1}=+P+89-T_{AD}$$

Particle A move with same acceleration upward, Newton second law:
$$-9,08.a_{1}=+89-T_{AD}$$

Substituting the equations gives me: $$2.T_{AD}=178+P$$

Now I do have to find a relation between the acceleration of the subsystem and particle B.
I know that the effective acceleration of particle B is:$$1,22 m/s^{2}$$.
But I don't understand why $$a_{1}$$ equals $$9,81 m/s^{2}-1,22m/s^{2}$$

Could you explain this to me?

4. Mar 28, 2012

### Staff: Mentor

I don't think I can explain it, because I don't think that it's true if I'm correctly understanding that "9,81" is g and "1,22" is the net acceleration of block B

I believe that the acceleration a1 of the subsystem as a whole (and hence block A also) should be closer to 0.4 m/s2.

What sort of "relation between the acceleration of the subsystem and particle B" are you looking for and why? Didn't already solve for force P?

5. Mar 28, 2012

### HWGXX7

Here are my usefull equation to solve problem:
$$2.T_{AD}=178+P$$
$$9,08.a_{1}=P+89-T_{AD}$$
The acceleration of particle B consist of 2 components:
$$a_{1}=a_{pully}$$
$$a_{Beffectif}=1,22 m/s^{2}=a_{1}+a_{particle B}$$
$$m_{particle B}.a_{particle B}=P+44,5-T_{BC}$$

Particle C will accelerat upwards because of force P and the subsystem will move down.

With this configuration I still cannot solve it..

By the way the solution you proposed is indeed correct.

grtz

6. Mar 31, 2012

### HWGXX7

Oke, I think I got the correct appoarch with the relevant assumptions according to the kinematics. It was actualy a bit more tricky than just putting in some variabeles.

I will write the solution-equation in the assumption the system is moving:

- particle A: $$-m_{A}.a_{A}=F_{g,A}-T_{AD}$$
- subsystem (containing pully D en particle B and C): $$+(m_{B}+m_{C}).a_{A}-T_{AD}+P$$

This lead to: $$178-2.T_{AD}+P=0$$
$$18,14.a_{A}=P$$

Now the tricky part I first looked over: the given kinematic expression of particle B is abolute. This means that: $$1,44m/s^{2}=a_{B/D}+a_{D}$$

For particle C:$$a_{C}=a_{D}-a_{C/D}$$
$$a_{C/D}=a_{D}-a_{C}$$

We know that: $$a_{C/D}=a_{B/D}$$
The dynamics of particle C: $$a_{C}.m_{C}=+44,5-T_{BC}$$
$$T_{BC}=44,5+m_{A}.(a_{D}-a_{C/D})$$
So now I can write rope force BC as function of acceleration:
$$T_{BC}=44,5+m_{A}.(a_{D}-a_{A/D})$$

The dynamics of particel B: $$m_{B}.a_{B}=+P+44,5-T_{BC}=6,5N$$
$$m_{B}.a_{B}=+18,14.a_{A}+44,5-T_{BC}=6,5N$$
$$m_{B}.a_{B}=+18,14.(1,44-a_{B/D})+44,5-T_{BC}=6,5N$$

Putting this all together get's me: $$a_{D}=0,479 m/s^{2}$$

Still a bit incorrect, but more important to me is the method. So my final question are the assumptions I used correct?

grtz

7. Mar 31, 2012

### Staff: Mentor

Your assumptions and approach look okay to me.

Personally, I like to leave everything in symbolic form for as long as possible in order to avoid having to "carry around" a lot of numbers through a derivation and to avoid accumulating rounding errors.

8. Apr 1, 2012

### HWGXX7

Indeed, but I wanted to make it clear that I got the good approach. So with a couple of number it's more easily.

Offcourse the errors occured at my final solution is due to the fact of rounding solution at the steps I took.

thank for all the help!