To leave Nicholson, you exit via the door to the quad. The door is 40 inches wid

Question

To leave Nicholson, you exit via the door to the quad. The door is 40 inches wide and has a mass of 200 kg. You push the door with a perpendicular force of 49.06N exactly 0.91m away from the hinge. This generates the minimum torque, T1 , required to open the door.
If you pushed the door 0.24m away from the hinge, with what force (in N) must you push the door to supply the same torque? To leave Nicholson, you exit via the door to the quad. The door is 40 inches wide and has a mass of 200 kg. You push the door with a perpendicular force of 49.06N exactly 0.91m away from the hinge. This generates the minimum torque, T1 , required to open the door.
If you pushed the door 0.24m away from the hinge, with what force (in N) must you push the door to supply the same torque? To leave Nicholson, you exit via the door to the quad. The door is 40 inches wide and has a mass of 200 kg. You push the door with a perpendicular force of 49.06N exactly 0.91m away from the hinge. This generates the minimum torque, T1 , required to open the door.
If you pushed the door 0.24m away from the hinge, with what force (in N) must you push the door to supply the same torque?

Explanation / Answer

We know that the equation of torque is given by

Torque = (Force) x (Distance)

Let us consider torque T1 required to open the door.

T1 = (49.06 N) * (0.91 m)

Let us consider torque T2required to close the door.

T2 = (Force) * (0.24 m)

Equating the torque in both cases, we get

T1 = T2

(49.06 N) * (0.91 m) = (Force) * (0.24 m)

44.64 = (Force) * (0.24)

(Force) = (44.64) / (0.24)

Force = 186 N

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