# Laws of motion question with constraint relations

1. Oct 16, 2011

### capricornpeer

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

Find :
a) acceleration of 1 kg, 2kg and 3 kg blocks and
b) tensions T1 and T2

Note: In the figure, 1, 2 and 3 represent the masses of respective blocks in Kg. T1 and T2 represent tension in strings

2. Relevant equations
Newton's laws and constraint equations

3. The attempt at a solution
Considering the uppermost pulley as a reference point, I have taken distance to :
i) mass 1 as x1,
ii) mass 2 as x2,
iii) mass 3 as x3.

From constraint considerations, T1=T2= T(say)

keeping string lengths as constant gives a1+a2+a3=0 (after a bit of solving).

Final Equations :
1) constraint equation : a1+a2+a3=0
2) for block 1 : T = -a1
3) for block 2 : 2*g - T = 2*a2
4) for block 3 : 3*g - T = 3*a3

However, the problem is that ans given in the book (DC Pandey, Mechanics Vol.1) doesn't

Answer given in the book : a1 = 560/51, a2 = 590/51, a3 = 530/51 , T = 560/51 (all in SI units)

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

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### abc.jpg
File size:
15 KB
Views:
285
2. Oct 17, 2011

### ehild

There are two independent strings. Why should be the tension equal in them? It can be, but you should prove.

The whole system accelerates downward, why is the sum of the accelerations zero? It can be if you choose the direction of accelerations this way, but why do you say that a1 is opposite to the tension?
Show the direction of the accelerations.

Find relation between accelerations using the constraints for both strings.

ehild

Last edited: Oct 17, 2011
3. Oct 23, 2011

### capricornpeer

I had proved that tensions are equal and just stated the result on the forum. I think that's pretty trivial to see. Apologies for that.

I have chosen the uppermost pulley as reference for displacements x1,x2 and x3. Now, the vector x1's tail is at this pulley and it points towards block 1. Therefore, the direction of a1 is horizontal and pointing towards block 1. Also, since the block 1 will move towards right, a1 will be negative and thus a2 and a3 should be positive. Thus, sum of accelerations can be zero.

Please can you find the constraint relation between accelerations and check if it is indeed
a1+a2+a3=0?

Thanks

4. Oct 23, 2011

### Staff: Mentor

These accelerations are greater than g. They cannot be correct.

5. Oct 23, 2011

### ehild

Block 1 accelerates on the right. You can call the acceleration of block 1 by -a1, but a1 is not its acceleration then.
You need to use one coordinate system for a system of masses. Block1 is on the other side of the upper pulley than the other two blocks, so x1 equals to the negative of the distance from the pulley. The acceleration is not the second derivative of distance from the origin, but the second derivative of the position vector.

By the way, Doc Al is right, the given accelerations can not be true.

ehild

Last edited: Oct 23, 2011
6. Oct 24, 2011

### capricornpeer

Glad you pointed that out sir. Thanks

Actually, I took the acceleration of block 1 as a1 and wrote the following equations (the same as my first post) :
1) a1+a2+a3=0
2) for block 1 : T = (-)*(a1) (notice the minus sign)
3) for block 2 : 2*g - T = 2*a2
4) for block 3 : 3*g - T = 3*a3
Ans : a1=-1.09, a2=0.45, a3=0.64

If I take it as -a1, equations are :
1) -a1+a2+a3=0
2) for block 1 : T = (-)*(-a1) = a1
3) for block 2 : 2*g - T = 2*a2
4) for block 3 : 3*g - T = 3*a3
Ans : a1=1.09, a2=0.45, a3=0.64

Evidently, acceleration of block 1 can be taken as a1 and the negative sign in the answer indicates that the acceleration is in the direction opposite to that taken. The important thing to be kept in mind for both cases is: for block 1, T= -a1(in case 1) and T= -(-a1) (in case 2) as acceleration is known to be in the opposite direction from that assumed. Thanks anyways for replying.

7. Oct 24, 2011

### ehild

Do not forget Newton's second law: F=ma. The force is T, and it points on the right. So the acceleration of block 1 also points on the right. If you write T=-a1*m1, then a1 is not the acceleration of block 1. It is something else: Second derivative of the length of the piece of rope. If you keep saying that a1 is the acceleration and write ma1=-F it is against Physics.

ehild