You need to open a heavy door that's 40 inches wide. Instinctively, you push near the edge that's the farthest from the hinges. If instead, you had pushed at a point only 10 inches in from the hinges, how much harder would you have had to push to open the door at the same speed? explain your answer.

Question

You need to open a heavy door that's 40 inches wide. Instinctively, you push near the edge that's the farthest from the hinges. If instead, you had pushed at a point only 10 inches in from the hinges, how much harder would you have had to push to open the door at the same speed? explain your answer.

Answer 1

Four times more force are needed at ten inches from the hinge axis to produce the torque needed to open the door.

Answer 2

what is the equation to do this? help! please. lost. don't understand how you did this

Answer 3

To understand how much harder you would have to push to open the door at the same speed, we can consider the principles of torque. Torque is the rotational equivalent of force and is responsible for the turning motion of an object.

When you push a heavy door near the edge farthest from the hinges, you are exerting a larger force on the door compared to when you push it at a point closer to the hinges. This is because the distance between the point of application of the force (where you push) and the axis of rotation (the hinges) affects the torque.

The torque acting on the door is given by the equation:

Torque = force x distance

In this case, we can assume that the torque required to open the door at the same speed is constant, irrespective of the point of application. So, let's compare the torques for both cases.

Case 1: Pushing near the edge farthest from the hinges (originally)

Assuming you exert a force F1, which is sufficient to open the door at the desired speed, the torque exerted would be:

Torque1 = F1 x D1

Here, D1 is the distance between the point of application of the force (where you push) and the hinges.

Case 2: Pushing at a point only 10 inches in from the hinges (alternate scenario)

Now, if you push at a point only 10 inches in from the hinges, the torque exerted would be:

Torque2 = F2 x D2

Here, D2 is the new distance between the point of application of force and the hinges (10 inches in this case). Note that we don't know the exact value of F2.

Since we want to open the door at the same speed in both scenarios, the torque exerted in Case 1 (Torque1) should be equal to the torque exerted in Case 2 (Torque2). Mathematically, we can express it as:

F1 x D1 = F2 x D2

Now, we can determine how much harder you would have to push (F2) in the second case compared to the first (F1). Rearranging the equation, we get:

F2 = (F1 x D1) / D2

Plug in the known values:

F2 = (F1 x 40 inches) / 10 inches

F2 = 4 x F1

From this calculation, we can conclude that you would have to push four times harder (F2 = 4 x F1) in the second scenario, where you push near the hinges, to achieve the same speed of opening the door as when you push near the edge farthest from the hinges.