# Find masses of particles on pulleys attached by strings

1. Oct 8, 2015

### moenste

1. The problem statement, all variables and given/known data
A particle A is suspended as shown in the figure by two strings, which pass over smooth pulleys, and are attached to particles B and C. At A, the strings are at right angles to each other, and make equal angles with the horizontal. If the mass of B is 1 kg, what are the masses of A and C?

Answers: A = √2 kg (20.5 kg), C = 1 kg

2. The attempt at a solution
For C:
If B = 1 kg and then it should be equal to C and C = 1 kg. I would say that because the system is in equilibrium and this is the reason, but it is not stated that the system is in eq. so I don't quite know why C is equal to 1 kg and not 1.1 or 0.9 kg.

For A:
At first I thought that A = B + C so 2 kg which is wrong. Then I though that if we take B as 1 kg, the angle as 45 degrees then A should be 1 cos 45. Adding another 1 cos 45 due to C is 2 cos 45 = 1.41 kg or root 2. But again, not sure about whether this logic is correct.

2. Oct 8, 2015

### BvU

Well observed. Problem isn't solvable if not in eq, so you can safely assume it is.

Remember the conditions for A in eq ? there are two components to $\Sigma \vec F$ !

3. Oct 8, 2015

### moenste

In that case if we assume that the system is in eq. then C = B = 1 kg.

Concerning A: a body is in equilibrium if the resultant force on its centre of mass is zero, and the total torque about all axes is zero; condition: the resultant force in any two directions in the plane of the forces is zero. 1 cos 45 [ B] = 1 cos 45 [C] as I understand it.

4. Oct 8, 2015

### BvU

Yes. No worry about torque needed here. Just a simple sum of forces. Two tensions and one from gravity. The tensions in each of the strings are equal at each end (thanks to the pulleys). Directions are given.

5. Oct 8, 2015

### moenste

So, the solution as I understand:
1. First we assume that the system is in an equilibrium.
2. Because the system is in eq. then B = C = 1 kg.
3.
Tension in vertical lines is 10 N. Lines which are attached to A are also equal to 10 N (true? because they are the same vertical lines that is why they are 10 N?).
4. Drawing two triangles which have two 45 angles. We need to find the adjacent side in the triangle: 10 cos 45. Because we have two triangles then 2 * 10 cos 45 = 14.14 N. Diving by 10 m/s2 we get 1.41 kg or 20.5 = A.

This is correct?

6. Oct 8, 2015

### BvU

Very much so.

You still seem a bit insecure about these tensions on the wires: If the tensions were not equal, the wire itself would accelerate!

The fact that the system is in equilibrium is not enough to establish mB = mC , though. Can you think of the deciding information in the problem statement that does this ?

7. Oct 9, 2015

### moenste

I also read in a different source that due to "smooth pulleys" there is no tension and that's why 10 N is thoughout the whole wire. This logic is correct?

As I think it's either "smooth pulleys", so there is no tension and 10 N of the B side should be equal to 10 N of the C side and -> C = 1 kg. Or it's because the angle is 90 degrees at A so the masses of both B and C should be equal.

8. Oct 12, 2015

### BvU

A smooth pulley can not exercise a force along the wire anywhere where it touches the pulley, so it can't change the tension: there's no tension difference between the two ends of a wire over a smooth pulley.
If a pulley is not smooth, there still are several possibilities: the wire can slip or not and the pulley rotation can be free or with friction. For a static problem the slipping isn't an issue, but friction in the pulley bearing can exercise a torque on the wire -- which gives a tension difference between the ends of the wire.

No friction is indeed a necessary condition. The 90 degrees actually isn't. Together, that's still not enough. The "the strings make equal angles with the horizontal" is essential too and clinches it: The main and crucial argument for $m_B = m_C$ is symmetry: your pulleys are static as well, so they both must be in equilibrium, with $\ \ \Sigma \vec F = 0\ \$. Since they are frictionless, the tensions in the vertical direction are the same as in the 45 degree direction. For A too you have $\ \ \Sigma \vec F = 0\ \$, so the horizontal components of the tension forces on A must be opposite and equal, which forces $m_Bg\cos 45^\circ = m_Cg\cos 45^\circ$ .

You can check for yourself if you now understand the whole thing: repeat the exercise with 60$^\circ$ or 120$^\circ$ instead of the 90$^\circ$

9. Oct 12, 2015

### moenste

So, for 60 degrees it's: mB g cos 30 = mC g cos 30 = 1 kg and 2 kg * g * cos 30 = 17.32 N and A = 1.73 kg. For 120 degrees it's: mB g cos 60 = mC g cos 60 = 1 kg and 2 kg * g * cos 60 = 10 N and A = 1 kg.

10. Oct 12, 2015

Spot on !