An uniform 8m metal rod AB of weight 400N is hinged to the wall at point B. The end A of the rod is connected with a string of length 6m to the wall at point C.
Calculate the tension of the string.

The given problem involves a uniform rod \( AB \) and a string attached to the wall at point \( C \). This is a typical static equilibrium problem where we need to calculate the tension in the string.

### Assumptions:
- The rod \( AB \) is hinged at point \( B \) and is kept in equilibrium by the string \( AC \).
- The rod is uniform, so its weight acts at its center of mass, which is at the midpoint of the rod.
- The tension in the string and the reactions at the hinge at \( B \) keep the system in equilibrium.

Let's break the problem into steps and calculate the tension in the string.

### Step 1: Label the known values.
- Length of the rod \( AB = 8 \, \text{m} \)
- Weight of the rod \( W = 400 \, \text{N} \)
- Length of the string \( AC = 6 \, \text{m} \)

### Step 2: Geometrical considerations.
- Let’s assume the point \( B \) is on the wall vertically below point \( C \).
- The horizontal distance between points \( B \) and \( C \), denoted by \( BC \), is 6 m (since the string is fully stretched and horizontal).
- The rod makes an angle with the horizontal, and we can calculate it using geometry.

### Step 3: Torque equilibrium.
The system is in static equilibrium, so the sum of torques about any point must be zero. Let's choose point \( B \) as the pivot point to eliminate the reaction forces at the hinge.

The two forces causing torque are:
1. The weight of the rod acting at its center of mass (at 4 m from point \( B \)).
2. The tension in the string acting at point \( A \).

#### Torque due to the weight of the rod:
The weight \( W = 400 \, \text{N} \) acts at the midpoint of the rod (4 m from point \( B \)). The torque due to this force is:
\[
\tau_{\text{weight}} = W \times \text{perpendicular distance from } B
\]

The perpendicular distance is calculated using the angle between the rod and the horizontal, which we can find from the geometry.

Let’s calculate the angle of the rod first. The right triangle formed by \( A \), \( B \), and \( C \) has:
- \( AB = 8 \, \text{m} \) (rod length)
- \( BC = 6 \, \text{m} \) (horizontal distance)

So, using Pythagoras' theorem:
\[
AC = \sqrt{AB^2 - BC^2} = \sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} \approx 5.29 \, \text{m}
\]


An uniform 8m metal rod AB of weight 400N is hinged to the wall at point B. The end A of the rod is connected with a string of length 6m to the wall at point C. Calculate the tension of the string.

Learn how to calculate the tension in a string attached to a uniform 8m rod using principles of static equilibrium and torque. This problem involves Pythagoras' theorem and the principle of moments, making it suitable for students in Grade 11-12 physics and advanced mathematics.

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