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 F1F0?
To determine the relationship between the forces F1 and F0, we can make use of the principles of torque and rotational motion.
First, let's consider the torque exerted on the door when it is pushed at a horizontal distance x0 from the hinge. The torque τ is given by the equation:
τ = Fr
Where F is the applied force and r is the distance from the point of rotation (hinge) to where the force is applied (x0 in this case). Therefore, the torque exerted at this distance is given by:
τ0 = F0 * x0
Next, let's consider the torque exerted on the door when it is pushed at a horizontal distance 2x0 from the hinge. The torque τ1 is given by:
τ1 = F1 * (2x0)
Since the angular acceleration α is the same in both cases, we can equate the torques and solve for F1:
F0 * x0 = F1 * (2x0)
Simplifying the equation:
F1 = F0/2
Therefore, the ratio of F1 to F0 is:
F1/F0 = (F0/2) / F0 = 1/2
So, F1F0 = (1/2) * 1 = 1/2.
In conclusion, F1F0 is equal to 1/2.