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

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Answer 2

The second law for rotation

M=Iα => α=M/I
M₁/I₁=M₀/I₀ =>
M₁/M₀=I₁/I₀
I₁ = I₀ =>
M₁ = M₀
F₁x₁=F₀x₀
F₁/F₀=x₀/x₁=x₀/2x₀ =1/2

Answer 3

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Answer 4

To determine the relationship between F1 and F0, we can use the concept of torque. Torque is the rotational equivalent of force and is given by the equation:

Torque (τ) = Force (F) × Distance (r)

In this case, the torque required to open the door is proportional to the force applied and the distance from the hinge.

Let's assume the door has a uniform mass distribution and rotates around a fixed hinge. The torque required to accelerate the door is given by:

τ = Moment of Inertia (I) × Angular Acceleration (α)

The moment of inertia depends on the shape and mass distribution of the door. However, for simplicity, let's assume the moment of inertia (I) is constant.

Now, let's consider the two situations:

Situation 1: Pushing at a distance x0

In this case, the torque τ1 required to open the door is given by:
τ1 = F0 × x0

Situation 2: Pushing at a distance 2x0

Similarly, the torque τ2 required to open the door is given by:
τ2 = F1 × 2x0

Since both situations have the same angular acceleration α, we can equate the torques:

τ1 = τ2
F0 × x0 = F1 × 2x0

Now, let's solve for F1 in terms of F0:

F1 = (F0 × x0) / (2x0)
F1 = F0 / 2

Therefore, the relationship between F1 and F0 is F1 = F0 / 2.

So, F1 (the amount of force required to open the door at 2x0 distance) is half the amount of force, F0, required to open the door at x0 distance.

Therefore, F1/F0 = (F0 / 2) / F0 = 1/2.

In conclusion, you would need half the amount of force to open the door when pushing perpendicularly at a horizontal distance 2x0 from the hinge compared to pushing at a distance x0.