To calculate the maximum torque that the crane can withstand, we can use the formula for torque:

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

Where:

• $\tau$ is the torque,
• $r$ is the length of the crane's arm (15.0 m),
• $F$ is the maximum load force the crane can handle (799 N),
• $\theta$ is the angle between the arm and the horizontal (20.0°),
• $\sin(\theta)$ is the sine of the angle.

Let's plug in the values:

$\tau = 15.0 \, \text{m} \cdot 799 \, \text{N} \cdot \sin(20.0^\circ)$

First, we calculate $\sin(20.0^\circ)$: $\sin(20.0^\circ) \approx 0.3420$

Now substitute that back into the torque equation: $\tau = 15.0 \, \text{m} \cdot 799 \, \text{N} \cdot 0.3420$ $\tau \approx 4095.99 \, \text{N} \cdot \text{m}$

So, the maximum torque the crane can withstand is approximately:

$\tau \approx 4096 \, \text{N} \cdot \text{m}$

Would you like further details on torque calculation or its applications?

Here are 5 related questions to explore:

1. How does torque depend on the length of the arm in a crane?
2. What happens to the torque if the angle is increased?
3. How does the sine function affect torque in different scenarios?
4. Can the crane handle the same torque if the load is applied at a different angle?
5. How does the direction of the applied force impact torque?

Tip: The sine of an angle peaks at 90°, meaning torque is maximized when the force is applied perpendicular to the arm.