# Video: Finding the Torque on an Object given a Linear Force Applied to It and the Distance to the Center of Rotation

When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850 m from the hinges. What torque are you exerting relative to the hinges?

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### Video Transcript

When opening a door, you push on it perpendicularly with a force of 55.0 newtons at a distance of 0.850 meters from the hinges. What torque are you exerting relative to the hinges?

The force applied we’re told is 55.0 newtons; we’ll call that 𝐹. We’re also told that this force is applied at distance of 0.850 meters from the hinges; we’ll call that distance 𝑟. If we were to draw a diagram, looking down on the door from above of the action going on, we would see that we’re applying the force 𝐹 perpendicularly to the door a distance of 𝑟 meters away from the door’s hinge.

We want to solve for the torque that we’re exerting relative to the hinges, which we’ll call 𝜏. To begin, we can recall the definition for torque. In general, the torque is equal to the cross-product of the radius 𝑟 with the applied Force 𝐹.

When 𝑟 and 𝐹 are in the same plane as they are in our example, we can write a simplified version of this equation for torque. Torque is equal to 𝑟 times 𝐹 times the sin of the angle between them. In our case, 𝑟 and 𝐹 are at right angles to one another. So when we use this simplified equation for torque, since the sin of 90 degrees is equal to one, the equation simplifies to simply torque equals 𝑟 cross 𝐹.

When we plug in for the given values of 𝑟 and 𝐹 and multiply them together, we find that the resulting torque is 46.8 newton meters. This is how much torque the force 𝐹 exerts relative to the hinges.