If you watch little kids try to open doors, they will often just push anywhere on the door, rather than on the edge like adults do. This is because they don't understand that it's easier to open a door if you push on the edge. But how much easier? Consider the following situation: you push on a door perpendicularly at a horizontal distance x0 from the hinge with a force F0, thereby opening the door with some angular acceleration α. Let F1 be the amount of force you'd need to exert to open the door with the same angular acceleration, but pushing perpendicularly at a horizontal distance 2x0 from the hinge. What is F1/F0?

Question

If you watch little kids try to open doors, they will often just push anywhere on the door, rather than on the edge like adults do. This is because they don't understand that it's easier to open a door if you push on the edge. But how much easier? Consider the following situation: you push on a door perpendicularly at a horizontal distance x0 from the hinge with a force F0, thereby opening the door with some angular acceleration α. Let F1 be the amount of force you'd need to exert to open the door with the same angular acceleration, but pushing perpendicularly at a horizontal distance 2x0 from the hinge. What is F1/F0?

Answer 1

To solve this problem, we can use the concept of torque. Torque is defined as the product of force and the perpendicular distance from the pivot point (hinge) to the line of action of the force.

In this case, we want to compare the force required to open the door at distance x0 versus the force required at distance 2x0. Let's denote the torque required at distance x0 as τ0 and the torque required at distance 2x0 as τ1.

We can use the following equation to calculate torque:
τ = r * F * sin(theta)

Here, r represents the distance from the pivot point to the line of action of the force, F represents the force applied perpendicular to the door, and theta represents the angle between the force and the line connecting the pivot point and the point of application of the force.

Since the force is applied perpendicularly in this scenario, the angle theta is 90 degrees, and the equation simplifies to:
τ = r * F

For distance x0, the torque τ0 is given by:
τ0 = x0 * F0

For distance 2x0, the torque τ1 is given by:
τ1 = 2x0 * F1

Since we are comparing the same angular acceleration for both cases, we can equate the torques:
τ0 = τ1

x0 * F0 = 2x0 * F1

Now we can solve for F1:
F1 = (x0 * F0) / (2x0)

Simplifying the equation:
F1 = F0 / 2

Finally, we can determine the ratio F1/F0:
F1/F0 = (F0 / 2) / F0 = 1/2

Therefore, F1/F0 is equal to 1/2. This means that to achieve the same angular acceleration, you only need half the force when pushing at a distance 2x0 from the hinge compared to pushing at distance x0.