A uniform rod of length L and density rho is being pulled along a smooth floor with a horizontal acc

Q: Why did we choose the rear half of the rod for our FBD instead of the front half?

You can choose either half! However, analyzing the rear half is simpler because the only horizontal force acting on it is the internal tension $T$ at the cut. If you analyzed the front half, you would have to include both the external pulling force at the front end and the internal tension $T$ at the cut, which requires more algebraic steps but yields the exact same result.

Q: Does the cross-sectional area of the rod affect the stress at the mid-point?

No, it does not. Although a larger area increases the mass of the rod (requiring more force to accelerate), it also distributes that force over a larger area. These two effects cancel each other out perfectly, making the stress independent of the cross-sectional area $A$.

Q: What would be the stress at the very front end of the rod?

At the very front end, the tension is equal to the total pulling force required to accelerate the entire rod, which is $T_{\text{max}} = M\alpha = (\rho A L)\alpha$. Therefore, the stress at the front end is maximum and is equal to $\sigma_{\text{max}} = \rho L \alpha$.

Q: Why is the stress at the rear end of the rod zero?

At the rear end, there is no mass behind it that needs to be pulled or accelerated. Since no force is required to accelerate a zero-mass section, the internal tension and consequently the stress at the free rear end are both zero.

Q: How does the stress vary along the length of the rod?

The stress varies linearly along the length of the rod. At a distance $x$ from the rear end, the stress is given by $\sigma(x) = \rho x \alpha$. It starts at $0$ at the rear end ($x=0$) and increases linearly to $\rho L \alpha$ at the front end ($x=L$).

Question

A uniform rod of length $L$ and density $\rho$ is being pulled along a smooth floor with a horizontal acceleration $\alpha$ (see figure). The magnitude of the stress at the transverse cross-section through the mid-point of the rod is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a uniform rod of length $L$, cross-sectional area $A$, and density $ho$ resting on a **perfectly smooth horizontal floor**.

The rod is being pulled to the right with a constant horizontal acceleration $\alpha$. Because the floor is smooth, there are no frictional forces opposing this motion.

Our goal is to find the **internal stress** developed at the transverse cross-section passing through the **mid-point** of the rod.

### The Master Equation

To begin, let us recall the definition of mechanical stress. **Stress** ($\sigma$) at any cross-section is defined as the internal restoring force ($T$) developed per unit cross-sectional area ($A$):

$$\sigma = \frac{T}{A}$$

Here, $T$ is the internal tension at the mid-point of the rod. To find $\sigma$, we must first determine the value of $T$.

### Isolating the Rear Half

To find the internal tension $T$, we can use the **method of sections**. Let us mentally cut the rod into two equal halves at its mid-point.

Each half has a length of $\frac{L}{2}$. Let us isolate the rear half of the rod and analyze its motion.

The only horizontal force acting on this rear half is the **tension force** $T$ pulling it forward, exerted by the front half of the rod at the interface.

### Calculating Mass and Force

The mass ($m$) of this rear half is given by the product of its volume and density:

$$m = 	ext{Volume} 	imes 	ext{Density} = \left(A \cdot \frac{L}{2}ight) ho$$

Since this rear half is accelerating to the right with the same acceleration $\alpha$ as the entire rod, we can apply **Newton's Second Law of Motion** ($F = ma$):

$$T = m \alpha$$

Substituting the value of mass $m$ into this equation:

$$T = \left(ho A \frac{L}{2}ight) \alpha$$

### Final Calculation of Stress

Now that we have the internal tension $T$, we can substitute it back into our master equation for stress:

$$\sigma = \frac{T}{A} = \frac{ho A \frac{L}{2} \alpha}{A}$$

Notice that the cross-sectional area $A$ cancels out beautifully from both the numerator and the denominator. This yields the final expression for the stress at the mid-point:

$$\sigma = \frac{1}{2} ho L \alpha$$

This elegant result shows that the stress at the mid-point depends only on the density of the material, the length of the rod, and its acceleration.

Visualized Solution

Visualizing the Accelerating Rod

Defining Stress at the Mid-point

Isolating the Rear Half

Mass of the Rear Half

Applying Newton's Second Law

Calculating the Mid-point Stress

Stress Distribution Along the Rod

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

The Master Equation

Isolating the Rear Half

Calculating Mass and Force

Final Calculation of Stress

Similar Questions

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A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$ . If the wire is stretched by an amount $x$ , the work done is ......

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A steel wire having a radius of $2.0 mm$ , carrying a load of $4 kg$ , is hanging from a ceiling. Given that $g = 3.1 π ms^{- 2}$ , what will be the tensile stress that would be developed in the wire?

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Visualized Solution

Visualizing the Accelerating Rod

Defining Stress at the Mid-point

Isolating the Rear Half

Mass of the Rear Half

Applying Newton's Second Law

Calculating the Mid-point Stress

Stress Distribution Along the Rod

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

The Master Equation

Isolating the Rear Half

Calculating Mass and Force

Final Calculation of Stress

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A wire of length L and cross-sectional area A is made of a material of Young's modulus Y. If the wire is stretched by an amount x, the work done is ......

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A wire fixed at the upper end stretches by length l by applying a force F. The work done in stretching is

A wire of length $L$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$ . If the wire is stretched by an amount $x$ , the work done is ......

A wire fixed at the upper end stretches by length $l$ by applying a force $F$ . The work done in stretching is