NLM O7

Solution

We use the method of similar triangles to analyze the forces on the metal ball.

The ball is in equilibrium under the action of three concurrent forces:

1. Tension force ($F$) acting along the string $PB$.
2. Normal reaction force ($R$) acting radially outward along the radius $OB$.
3. Weight ($mg$) acting vertically downwards, parallel to $PO$.

Observe the geometry triangle $\Delta POB$:

• Side $PB$ is along the Tension $F$.
• Side $OB$ is along the Normal $R$.
• Side $PO$ is along the vertical (parallel to $mg$).

Since the directions of the forces match the directions of the sides of $\Delta POB$, the triangle of forces is similar to the triangle $\Delta POB$.

We can write the ratio of force magnitude to side length:

$$ \frac{F}{PB} = \frac{R}{OB} = \frac{mg}{PO} $$

From the ratio, we can express $F$ and $R$:

$$ R = mg \left( \frac{OB}{PO} \right) $$ $$ F = mg \left( \frac{PB}{PO} \right) $$

Here:
$mg$, $OB$ (radius of hemisphere), and $PO$ (radius of hemisphere) are all constants.
Therefore, $R$ is constant (remains unchanged).

However, $PB$ represents the length of the string segment from the pulley to the ball. As the ball is pulled up towards the pulley, the length $PB$ decreases.
Since $F \propto PB$, as $PB$ decreases, $F$ decreases.