Painter on a Scaffold - Don't Fall Off!!

Summary

In this physics lesson, Mr. P demonstrates a static equilibrium problem involving a painter on a scaffold. The task is to determine the closest point a 200 g object (representing the painter) can be placed on a 93 g meter stick without unbalancing the system. Using concepts of torque and forces, the students calculate that the painter can stand 6.0 cm from the left end of the meter stick before the system becomes unstable. Through a hands-on demonstration, they confirm that the solution works, illustrating how changes in normal forces impact the balance of the system.

Takeaways

• 😀 The problem involves a painter on a scaffold, supported at two points, which is a typical static equilibrium problem.
• 😀 The scaffold is modeled using a meter stick, two dominos as supports, and a 200-gram object representing the painter.
• 😀 The task is to determine the closest the painter can be placed to the end of the meterstick while keeping the system balanced.
• 😀 The meterstick's mass is 93 grams, and it is supported at the 20.0 cm and 80.0 cm marks.
• 😀 The forces acting on the system include the gravitational force on the painter and the meterstick, and the normal forces from the supports.
• 😀 The system is in rotational equilibrium, meaning the sum of torques acting on it equals zero.
• 😀 The torque is calculated using force and distance (lever arm), with the center of mass of the meterstick located at the 50 cm mark.
• 😀 As the painter moves along the scaffold, the normal forces at the supports change, eventually reaching zero on one side.
• 😀 The system will become unbalanced if the painter moves too far toward one end, causing the scaffold to fall.
• 😀 The closest the painter can be to the left end of the meterstick while maintaining equilibrium is 6.0 cm.
• 😀 The final calculations reveal that the normal force at one of the supports reaches zero when the painter is 6.0 cm from the left end, marking the point where the system remains in balance.

Q & A

What is the main goal of the problem presented in the transcript?

The goal is to determine the closest point from the end of a meter stick where a 200g object (representing a painter) can be placed without unbalancing the system, assuming the meter stick is supported at two points (20.0 cm and 80.0 cm).

What are the key forces acting on the system in this problem?

The key forces include the gravitational force acting downward on the meter stick and the object, and the normal forces from the two supports (dominos), acting upward at the 20.0 cm and 80.0 cm marks.

Why does the torque caused by the normal force at the 20 cm mark not contribute to the rotation?

The torque caused by the normal force at the 20 cm mark does not contribute to the rotation because it acts directly at the axis of rotation, meaning the distance (r) from the pivot point is zero, and thus its torque is also zero.

What is the significance of using torques instead of summing forces to solve the problem?

Using torques allows for a simpler solution by focusing on rotational equilibrium. Summing forces is not always necessary because the forces in the vertical direction are balanced by the normal forces at the supports, so analyzing torques gives a more direct way to find the balance point.

What happens to the normal force on the right support as the painter moves closer to the left end?

As the painter moves closer to the left end, the normal force on the left support increases, while the normal force on the right support decreases. Eventually, the normal force on the right becomes zero, indicating that the system is on the verge of losing balance.

What condition must be met for the system to remain balanced?

For the system to remain balanced, the sum of the torques must be zero, meaning the torques produced by the object and the meter stick must be equal and opposite. This ensures that the system is in rotational equilibrium.

How is the distance from the left end of the meter stick calculated for the object?

The distance is calculated by finding the torque balance. The torque due to the object is set equal to the torque due to the meter stick, and the position of the object (r) is found to be 13.95 cm from the 20 cm mark. Using the equation x = 20 cm - r, the closest distance is determined to be 6.0 cm.

What is the final value for the closest distance the painter can be from the left end of the meter stick?

The final value for the closest distance the painter can be from the left end of the meter stick is 6.0 cm.

What does the removal of the domino in the experiment demonstrate?

The removal of the domino demonstrates that the system remains balanced even when one of the supports (the one with zero normal force) is no longer providing any support. This confirms that the calculations were correct and the system stays in equilibrium.

Why is the angle of 90 degrees important in the torque calculations?

The angle of 90 degrees is important because it ensures that the torque calculation is simplified. Since both forces of gravity are acting vertically and the distances (r) are horizontal, the sine of 90 degrees equals 1, making the torque calculations straightforward and eliminating the need for angle adjustments.