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 answer the question, we need to consider the principles of torque and rotational equilibrium. Torque is the measure of a force's ability to cause an object to rotate about an axis. In this case, the axis of rotation is the hinge of the door.

The torque applied to the door is given by the equation:

τ = r * F * sin(θ)

where:
τ is the torque,
r is the distance from the hinge (perpendicular to the force direction),
F is the applied force, and
θ is the angle between the applied force and the line connecting the hinge and the point of application.

In our case, the force is perpendicular to the door (θ = 90 degrees), and we want to compare the forces required at two different distances from the hinge.

Let's consider the first situation, where the force is applied at a distance x0 from the hinge. The torque applied can be expressed as:

τ0 = x0 * F0

Now, let's consider the second situation, where the force is applied at a distance 2x0 from the hinge. The torque applied is:

τ1 = 2x0 * F1

Since we want to compare the torques in these two situations, we can set them equal to each other:

x0 * F0 = 2x0 * F1

We can simplify this equation by canceling out the x0 term:

F0 = 2F1

Therefore, we see that F1 = (1/2) * F0.

To find F1/F0, we can substitute the value of F1 we found into the equation:

F1/F0 = ((1/2) * F0) / F0

Simplifying the expression, we find:

F1/F0 = 1/2

Therefore, F1 is half the magnitude of F0, indicating that you would need to exert half the force at 2x0 distance to produce the same angular acceleration as at the x0 distance.