A wire is suspended from the ceiling and stretched under the action of a weight $F$ suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight. (Note: more than one of the given options may be correct.) (a) Tensile stress at any cross section $A$ of the wire is $F/A$. (b) Tensile stress at any cross section is zero. (c) Tensile stress at any cross section $A$ of the wire is $2F/A$. (d) Tension at any cross section A of the wire is $F$.

> [!TLDR] > When a wire is suspended and stretched by a weight $F$, the internal pulling force (tension) at any cross-section is $F$. The tensile stress is this tension divided by the cross-sectional area $A$, resulting in $F/A$. > The correct options are **(A)** and **(D)**. To understand the forces and stresses within the wire, we need to define tension and tensile stress and then apply these definitions to the given scenario. **Understanding Tension and Tensile Stress** 1. **Tension:** Imagine a wire being pulled from both ends. The internal force that resists this pulling and transmits the force along the wire is called tension. If you were to conceptually cut the wire at any point, the force exerted by one part of the wire on the other part across that cut surface is the tension. For a massless wire in equilibrium, the tension is uniform throughout its length. 2. **Tensile Stress:** When a material is subjected to a tensile force (tension), it experiences internal forces distributed over its cross-sectional area. Tensile stress ($\sigma$) is defined as the internal restoring force per unit cross-sectional area. > [!FORMULA] > $\sigma = \frac{T}{A}$ > where $T$ is the tension (internal force) acting perpendicular to the cross-section, and $A$ is the cross-sectional area. **Step-by-Step Solution** 1. **Analyze the forces on the wire:** The wire is suspended from the ceiling. A weight $F$ is attached to its lower end, pulling it downwards. The problem states that the force exerted by the ceiling on the wire is equal and opposite to the weight, meaning the upward force from the ceiling is also $F$. This confirms the wire is in static equilibrium. 2. **Determine the tension at any cross-section:** Consider an arbitrary cross-section of the wire, say at a distance $x$ from the bottom end. * **Method 1: Considering the lower segment:** Imagine isolating the segment of the wire below this cross-section. This segment is in equilibriu

A wire is suspended from the ceiling and stretched under the action of a weight F suspended from its other end

NCERT Exemplar · Q10

F

(a) Tensile stress at any cross section

A

of the wire is

F/A

(b) Tensile stress at any cross section is zero.

A

of the wire is

2F/A

(d) Tension at any cross section A of the wire is

F

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Concept understanding — Static Equilibrium

Static Equilibrium: The Art of Staying Put

Imagine a book lying flat on a table. It doesn't move. A lamp hanging from the ceiling — still. A bridge holding up cars — motionless. What do all these have in common? They are in static equilibrium.

The word "static" means unchanging or stationary. "Equilibrium" comes from Latin aequilibrium — "equal balance." Put them together: a state where an object is completely at rest, with no tendency to start moving or rotating.

But here's the key insight: being at rest doesn't mean nothing is happening. Forces are still acting on that book — gravity pulls it down, the table pushes it up. The lamp feels tension from the ceiling pulling up and gravity pulling down. These forces are cancelling each other out perfectly.

Note

Static equilibrium is dynamic balance — forces are present, but their net effect is zero. The object "chooses" to stay still because all pushes and pulls are perfectly matched.

The Two Conditions for Static Equilibrium

For an object to be truly static (not moving or rotating), two separate things must be true simultaneously.

Condition 1: No Net Force (Translational Equilibrium)

The sum of all forces acting on the object must be zero. In vector form:

\sum \vec{F} = 0

This means:

All upward forces equal all downward forces
All leftward forces equal all rightward forces
All forward forces equal all backward forces

If you break it into components (the standard exam approach):

\sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0

Why this alone isn't enough: Imagine pushing a door at its handle — it rotates open. The forces might balance (you push, the hinges push back), but the door still moves. That's why we need the second condition.

Condition 2: No Net Torque (Rotational Equilibrium)

The sum of all torques (twisting effects) about any point must be zero:

\sum \tau = 0

Torque depends on three things: the force applied, the distance from the pivot point, and the angle at which you push. For a force $F$ applied at distance $r$ from the pivot, at angle $\theta$ :

\tau = rF\sin\theta

Watch out

A common mistake: thinking torque only matters if the object is actually rotating. Torque can be present even when nothing moves — it's just balanced by other torques. A seesaw with two kids of equal weight at equal distances is a perfect example.

Putting It All Together

For an object to be in static equilibrium:

\sum \vec{F} = 0 \quad \text{and} \quad \sum \tau = 0

Both conditions must hold simultaneously. If either fails, the object will either accelerate (move in a straight line) or start rotating (or both).

A Simple Example: The Book on the Table

Consider a 2 kg book on a horizontal table. Gravity pulls down with force $F_g = mg = 2 \times 9.8 = 19.6 \text{ N}$ .

The table pushes up with a normal force $N = 19.6 \text{ N}$ .

Check condition 1: $\sum F_y = N - F_g = 19.6 - 19.6 = 0$ ✓ …

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When a wire is suspended and stretched by a weight $F$ , the internal pulling force (tension) at any cross-section is $F$ . The tensile stress is this tension divided by the cross-sectional area $A$ , resulting in $F/A$ .

The correct options are (A) and (D).

To understand the forces and stresses within the wire, we need to define tension and tensile stress and then apply these definitions to the given scenario.

Understanding Tension and Tensile Stress

Tension: Imagine a wire being pulled from both ends. The internal force that resists this pulling and transmits the force along the wire is called tension. If you were to conceptually cut the wire at any point, the force exerted by one part of the wire on the other part across that cut surface is the tension. For a massless wire in equilibrium, the tension is uniform throughout its length.
Tensile Stress: When a material is subjected to a tensile force (tension), it experiences internal forces distributed over its cross-sectional area. Tensile stress ( $\sigma$ ) is defined as the internal restoring force per unit cross-sectional area.
$\sigma = \frac{T}{A}$

Step-by-Step Solution

Analyze the forces on the wire:

The wire is suspended from the ceiling. A weight $F$ is attached to its lower end, pulling it downwards. The problem states that the force exerted by the ceiling on the wire is equal and opposite to the weight, meaning the upward force from the ceiling is also $F$ . This confirms the wire is in static equilibrium.
Determine the tension at any cross-section:

Consider an arbitrary cross-section of the wire, say at a distance $x$ from the bottom end.
- Method 1: Considering the lower segment: Imagine isolating the segment of the wire below this cross-section. This segment is in equilibrium. The weight $F$ pulls it downwards. For it to be in equilibrium, the upper part of the wire (above the cross-section) must be pulling this lower segment upwards with a force equal to $F$ . This upward internal force is the tension $T$ at that cross-section. So, $T = F$ .
- Method 2: Considering the upper segment: Alternatively, consider the segment of the wire above the cross-section. The ceiling pulls this segment upwards with force $F$ . For this segment to be in equilibrium, the lower part of the wire (below the cross-section) must be pulling it downwards with a force equal to $F$ . This downward internal force is also the tension $T$ at that cross-section. So, $T = F$ . In both cases, the tension $T$ at any cross-section of the wire is equal to the suspended weight $F$ . …

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Q.A wire is suspended from the ceiling and stretched under the action of a weight FFF suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight. (Note: more than one of the given options may be correct.)

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Q.A wire is suspended from the ceiling and stretched under the action of a weight $F$ suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight. (Note: more than one of the given options may be correct.)

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