We have a ball of mass $m$ suspended by two unequal strings, where $T_1$ is longer than $T_2$. To determine the tensions $T_1$ and $T_2$, we analyze the forces acting on the ball.

Step 1: Define Forces

The forces acting on the ball are:

1. Tension forces $T_1$ and $T_2$, which act along the respective strings.
2. Gravitational force $mg$, which acts downward.

Since the ball is in equilibrium, the net force in both the horizontal and vertical directions must be zero.

Step 2: Set Up Equilibrium Equations

Let the angles that the strings make with the horizontal be $\theta_1$ (for $T_1$) and $\theta_2$ (for $T_2$). The force components in the horizontal and vertical directions are:

Horizontal equilibrium:

$T_1 \cos\theta_1 = T_2 \cos\theta_2$ $T_1 = T_2 \frac{\cos\theta_2}{\cos\theta_1}$

Vertical equilibrium:

$T_1 \sin\theta_1 + T_2 \sin\theta_2 = mg$

Step 3: Relationship Between Tensions and $mg$

• Since $T_1$ is associated with the longer string, its angle $\theta_1$ is smaller than $\theta_2$ (because a longer string tends to be more horizontal).
• Since $\cos\theta_1 > \cos\theta_2$, we get $T_1 < T_2$, meaning the shorter string supports more weight.
• The sum of the vertical components of both tensions must equal $mg$.

Key Relationships

• $T_1$ is smaller than $T_2$ (since the longer string has a smaller angle).
• The vertical components of $T_1$ and $T_2$ together balance the weight $mg$.
• The horizontal components of $T_1$ and $T_2$ must cancel each other.

Would you like a numerical example to illustrate this further?

Related Questions

1. How do the angles affect the distribution of tension?
2. What happens if one string is cut? Which way does the ball move?
3. How does the problem change if the strings are of equal length?
4. Can we solve this using a vector approach?
5. What if the ball is moving? How does dynamics come into play?

Tip:

In equilibrium problems, always break forces into components and apply Newton’s laws separately in each direction.