Hanging Spheres

Use the following diagram as a guide: $h$ stands for the height at which $A$ and $B$ are suspended (the middle string is horizontal, so it's the same for both). $T_{AB}$ is the tension in the string between $A$ and $B$ (same for both) and $T_A$ and $T_B$ are the tensions on the strings pulling diagonally upwards with angle $\alpha$ and $\beta$ on $A$ and $B$ respectively. The balls are in equilibrium, so the net force acting on each one of them must be zero. In particular, we can decompose the horizontal and vertical components of each ball:

1. $m_A \cdot g - T_A \cdot \sin \alpha = 0$
2. $T_{AB} - T_A \cdot \cos \alpha = 0$
3. $m_B \cdot g - T_B \cdot \sin \beta = 0$
4. $T_{AB} - T_B \cdot \cos \beta = 0$

By solving for $T_{AB}$ on equations 2 and 4, we get the following new equation:

$T_A \cos \alpha = T_B \cos \beta$

And by solving for $g$ on equations 1 and 3 we get the following one:

$\frac{T_A \sin \alpha}{m_A} = \frac{T_B \sin \beta}{m_B}$

Now we can solve for $T_B$ on the first new equation to get the following substitution:

$T_B = \frac{T_A \cos \alpha}{\cos \beta}$

And we can use this substitution on the other equation:

$\frac{T_A \sin \alpha}{m_A} = \frac{\frac{T_A \cos \alpha}{\cos \beta} \sin \beta}{m_B} = \frac{T_A \cos \alpha \sin \beta}{\cos \beta m_B}$

In the previous expression we can cancel out $T_A$ on both sides and rearrange the rest as so:

$\frac{m_B}{m_A} = \frac{\cos \alpha}{\sin \alpha} \cdot \frac{\sin \beta}{\cos \beta} = \frac{\tan \beta}{\tan \alpha}$

The tangent of $\alpha$ and $\beta$ can be obtained from the two right triangles on each side, having in mind that $\tan \theta = \frac{opposite}{adjacent}$ :

$\tan \alpha = \frac{h}{d_A}$

$\tan \beta = \frac{h}{d_B}$

By substitution in the previous equation we get:

$\frac{m_B}{m_A} = \frac{\tan \beta}{\tan \alpha} = \frac{h / d_B}{h / d_A} = \frac{d_A}{d_B}$

What we get from this last equation, namely $\boxed{\frac{m_B}{m_A} = \frac{d_A}{d_B}}$ , is that the mass of a ball is inversely proportional to its distance to the wall. Thus, the ball closer to the wall (in this case $A$ ) will have a bigger mass.

$\boxed{m_A > m_B}$

Suppose that the tensions in the slanting strings are $T_A,T_B$ , and that the tension in the horizontal string is $T$ . Suppose also that the slanting strings make angles $\theta_A,\theta_B$ with the horizontal. Since particle $A$ is in equilibrium, resolving forces horizontally and vertically gives $T \; = \; T_A\cos\theta_A \hspace{2cm} m_Ag \; = \; T_A\sin\theta_A$ Dividing these two equations gives $m_A \; = \; \frac{T}{g}\tan\theta_A$ Similarly, we obtain $m_B\; = \; \frac{T}{g}\tan\theta_B$ Since $d_A < d_B$ we have $\tan\theta_A > \tan\theta_B$ , and hence $\boxed{m_A > m_B}$ .

1. Since the balls are at rest, the horizontal component of tension in the leftmost string must equal the horizontal component of tension in the rightmost string.
2. This implies that vertical component of the tension in the leftmost string is $d_b/d_a$ times the vertical component of tension in the rightmost string.
3. This implies the ratio of the weight of ball A to the weight of ball B is $d_b/d_a$ .

Proof of 1: The balls are not accelerating so net horizontal force acting on them is zero.

Proof of 2: Follows from considering a right triangle with sides formed by the leftmost string, the left wall and a horizontal line connecting the center of a A to the left wall. Similarly consider also a triangle formed by the rightmost string, the right wall and a horizontal line connecting the center of B to the right wall. The tensions of the leftmost and rightmost strings, together with their horizontal and vertical components form triangles similar to the triangles just described. Together with statement #1, this implies statement #2.

Proof of 3: Follows from #2 and the fact that the weight of ball A equals the vertical tension in the leftmost string and the weight of ball B equals the vertical tension in the rightmost string.

Imagine you are holding both spheres, and have equal mass, if you release them, they will tend to go to the center, wich is different from the diagram, hence the ball A is heavier

Intuitively, if B were heavier than it is, then it would sink lower and cause the string A-B would tilt down to the right (no longer be horizontal). Also, adding weight to B would pull A further away from the left wall.

If weight were added to A, likewise B would be pulled further away from the right wall.

If A and B were equal in weight, and the corner mounted strings were the same length, then the string A-B would be horizontal. Increasing the length of the right corner-to-B string (as is the case in the diagram) would cause B to sink lower.

So we can see that a higher mass OR a longer string makes the ball sink lower. So if B has a longer string, then it must weigh less to compensate (and keep the connecting string horizontal).

Horizontal tension is constant because there are no horizontal forces excepts at the extremes of the rope. The vertical tension is equal to the weight of each mass, being the tract AB horizontal.

$\frac{m_A}{m_B}=\frac{N_A}{N_B}=\frac{\tan(\alpha_A)}{\tan(\alpha_B)}>1$

What does distance say about mass? Or how do they correlate? More energy is needed to propel more mass. So it's more likely that B would go further only if it weighed less than A which only traveled a smaller distance.

If you have to lift a ball on a string, like you are pushing a kid on a swing set, up past the point of equilibrium, that takes work. If you have to push the ball higher it takes even more work. As you can see from the diagram, B is "lifted higher" than A, which takes more work. So, if the balls are in equilibrium, yet evenly balanced, A must be doing more work than B, because it is harder to hold up B than it is A. So A is more massive.

Had it been Ma =Mb ,the combined cg would have been at centre but since da<db => cg towards Ma(for combined system of mass) => Ma>Mb

Let's take the case when D A = D B. By symmetry we can say that M A = M B. Now if we increase the mass of A, A goes down making the AB line slant. To make AB line horizontal, we have to pull the string connecting A to the wall. This makes D A < D B. Thus we obtain the case given in the question.

The moment of inertia of a sphere (I) = $\frac{2}{3}$ $MR^{2}$

A and B are identical. (Given)

I observe that since they are in equilibrium, the net torque (T) = 0.

$T_{A}$ = $T_{B}$

$\implies$ $I_{A}$ . $\alpha_{A}$ = $I_{B}$ . $\alpha_{B}$

$\implies$ $\frac{2}{3} M_{A}R^{2}$ . $\frac{a}{d_{A}}$ = $\frac{2}{3} M_{B}R^{2}$ . $\frac{a}{d_{B}}$

$\implies$ $\frac{M_{A}}{M_{B}}$ = $\frac{d_{A}}{d_{B}}$

Thus, if $d_{A} < d_{B}$ , then $M_{A} < M_{B}$