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?
To determine the ratio of F1 to F0, we can use the principle of moment, which states that the sum of the moments on an object must be zero for rotational equilibrium.
In this scenario, when pushing on a door, the moment arm is the perpendicular distance from the hinge to the line of action of the force. The moment arm in the first case (pushing at a distance x0) is x0, and in the second case (pushing at a distance 2x0), it is 2x0.
The moment (torque) τ produced by a force F about a pivot point can be calculated as τ = F * r, where F is the force applied perpendicular to the line connecting the point of application to the pivot point, and r is the moment arm.
Considering the first case, the moment (τ0) produced by force F0 is τ0 = F0 * x0. Since the door experiences an angular acceleration α, we have τ0 = I * α, where I is the moment of inertia of the door.
Now, let's consider the second case with force F1 and moment τ1 = F1 * (2x0). Since the angular acceleration is the same as in the first case (α), we have τ1 = I * α.
Since both moments are equal to I * α, we can equate the two equations:
τ0 = τ1
F0 * x0 = F1 * (2x0)
F1 = (F0 * x0) / (2x0)
F1 = F0 / 2
Therefore, F1 is half of F0. The ratio of F1 to F0 is F1 / F0 = 1/2.