Animated Solution for Physics - Laws of Motion: A rigid insulated wire frame in the form of a right angled triangle $ABC$, is set in a vertical plane as shown in figure. Two beads of equal masses $m$ each and carrying charges $q_1$ and $q_2$ are connected by a cord of length $l$ and can slide without friction on the wires. Considering the case when the beads are stationary determine (a) (i) The angle $\alpha$ (ii) The tension in the cord (iii) The normal reaction on the beads (b) If the cord is now cut what are the value of the charges for which the beads continue to remain stationary?

Imagine you are looking at a right-angled triangular wire frame, perfectly fixed in a vertical plane. Two beads, $P$ and $Q$, each of mass $m$, are threaded onto the wires $AB$ and $AC$ respectively. They are connected by a taut cord of length $l$. The beads carry charges $q_1$ and $q_2$, meaning they exert an electrostatic force $F_e$ on each other.

Our goal is to find the equilibrium conditions for this system. When the beads are stationary, the net force on each bead must be zero. Let's break down the forces acting on each bead. There are exactly three forces: 1. The downward gravitational force, $mg$. 2. The normal reaction from the wire, $N_P$ for bead $P$ and $N_Q$ for bead $Q$, acting perpendicular to the respective wires. 3. The forces acting along the cord: the mechanical tension $T$ pulling the beads together, and the electrostatic force $F_e$ pushing them apart (or pulling them together). We can elegantly combine these into a single effective force, $T-F_e$, acting along the line joining the beads.

Whenever an object is in equilibrium under the action of exactly three concurrent forces, Lami's theorem is a mathematical superpower. It states that the magnitude of each force is proportional to the sine of the angle between the other two forces.

To apply it, we must meticulously determine the angles between our three forces: $N$, $mg$, and $(T-F_e)$.

Let's establish our coordinate system. The horizontal base is $BC$. The wire $AB$ is tilted at $30^{\circ }$ to the horizontal. Therefore, the normal $N_P$, being perpendicular to $AB$, makes an angle of $90^{\circ }+30^{\circ }=120^{\circ }$ with the positive x-axis. The weight $mg$ points straight down at $270^{\circ }$.

The cord makes an angle $\alpha$ with the wire $AB$. Since $AB$ is at $30^{\circ }$, the cord is directed at an angle of $30^{\circ }-\alpha$ relative to the horizontal.

Let's calculate the angles between the forces for bead $P$: - Angle between $N_P$ ($120^{\circ }$) and $mg$ ($270^{\circ }$) is $270^{\circ }-120^{\circ }=150^{\circ }$. - Angle between $N_P$ ($120^{\circ }$) and the cord ($30^{\circ }-\alpha$) is $120^{\circ }-(30^{\circ }-\alpha )=90^{\circ }+\alpha$. - Angle between $mg$ ($270^{\circ }$) and the cord ($30^{\circ }-\alpha$) is $(360^{\circ }+30^{\circ }-\alpha )-270^{\circ }=120^{\circ }-\alpha$.

Applying Lami's theorem for bead $P$ yields our first master equation:

Now, let's shift our focus to bead $Q$ on wire $AC$. The wire $AC$ makes an angle of $180^{\circ }-60^{\circ }=120^{\circ }$ with the positive x-axis. The normal $N_Q$ is perpendicular to it, pointing at $120^{\circ }-90^{\circ }=30^{\circ }$.

The cord, viewed from $Q$ towards $P$, points in the opposite direction, at $180^{\circ }+(30^{\circ }-\alpha )=210^{\circ }-\alpha$.

Calculating the angles for bead $Q$: - Angle between $N_Q$ ($30^{\circ }$) and $mg$ ($270^{\circ }$) is $360^{\circ }-(270^{\circ }-30^{\circ })=120^{\circ }$. - Angle between $N_Q$ ($30^{\circ }$) and the cord ($210^{\circ }-\alpha$) is $210^{\circ }-\alpha -30^{\circ }=180^{\circ }-\alpha$. - Angle between $mg$ ($270^{\circ }$) and the cord ($210^{\circ }-\alpha$) is $270^{\circ }-(210^{\circ }-\alpha )=60^{\circ }+\alpha$.

Applying Lami's theorem for bead $Q$ gives our second master equation:

We now have a beautiful system of equations. Notice that both sets of equations contain the term $(T-F_e)$. Let's isolate it from both and equate them:

Using trigonometric identities, $\sin (90^{\circ }+\alpha )=\cos \alpha$ and $\sin (180^{\circ }-\alpha )=\sin \alpha$. The equation simplifies to:

Rearranging terms, we find the tangent of $\alpha$:

This immediately tells us that $\alpha =60^{\circ }$.

With $\alpha$ known, the rest of the puzzle falls into place. Let's substitute $\alpha =60^{\circ }$ back into our equation for $P$ to find the tension $T$:

Therefore, the tension is:

Similarly, we can find the normal reactions:

If the cord is suddenly cut, the mechanical tension $T$ instantly drops to zero. For the beads to remain magically suspended in their stationary positions, the equilibrium condition along the line connecting them must still hold true.

This means our effective force $(T-F_e)$ must still equal $mg$. Since $T=0$, we have:

This implies that the electrostatic force must be attractive (pulling the beads together) to counteract the component of gravity trying to slide them down the wires. Substituting Coulomb's law:

The negative sign confirms that the charges $q_1$ and $q_2$ must be of opposite polarity.