To find the gravitational torque acting on the beam about an axis through the hinged end, we can use the following formula:

$\tau = r \cdot F \cdot \sin(\theta)$

where:

• $\tau$ is the torque,
• $r$ is the distance from the hinge to the center of mass of the beam,
• $F$ is the force of gravity acting on the beam, and
• $\theta$ is the angle between the beam and the vertical.

Step 1: Identify the values

1. Length of the beam, $L$ = 2.0 m.
2. Mass of the beam, $m$ = 15 kg.
3. Gravitational acceleration, $g$ = 9.8 m/s².

Since the beam is uniform, its center of mass is at its midpoint, which is at a distance of $\frac{L}{2} = \frac{2.0}{2} = 1.0 \, \text{m}$ from the hinge.

The gravitational force acting on the beam is: $F = m \cdot g = 15 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 147 \, \text{N}$

Step 2: Determine the angle

In the middle position shown in the image, the beam is inclined at $45^\circ$ from the vertical. So, $\theta = 45^\circ$.

Step 3: Calculate the Torque

Now we can substitute the values into the torque formula:

$\tau = (1.0 \, \text{m}) \cdot (147 \, \text{N}) \cdot \sin(45^\circ)$

Since $\sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707$,

$\tau = 1.0 \cdot 147 \cdot 0.707 \approx 103.9 \, \text{N} \cdot \text{m}$

Answer

The gravitational torque on the beam about an axis through the hinged end at the middle position is approximately:

$\tau \approx 103.9 \, \text{N} \cdot \text{m}$

Would you like further details or have any questions?

Here are five related questions for practice:

1. How would the torque change if the angle was $30^\circ$ instead of $45^\circ$?
2. What would be the torque if the beam was horizontal?
3. How does the position of the center of mass affect the torque on the beam?
4. What would be the torque if the mass of the beam was doubled?
5. How would the gravitational torque differ if the hinge was at the midpoint of the beam?

Tip: Always check the angle's reference direction when calculating torque, as it affects the sine term.