If you push the outer edge of a 1.05-m-wide door with a 23.0-N tangential force, what torque results? What is the torque if you apply the same magnitude of force, in the same place, but at a 45 degree angle?
answer 1 = 23*1.05
abswer 2 = 23*1.05*cos 45
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To calculate the torque resulting from pushing the outer edge of a door, we need to know the perpendicular distance between the point of application of the force and the axis of rotation.
1. In the first scenario, where the force is applied tangentially:
The torque (τ) is given by the formula:
τ = r * F * sin(θ)
where r is the distance from the point of application of the force to the axis of rotation, F is the magnitude of the force, and θ is the angle between the force and the lever arm (which is 90 degrees in this case because the force is tangential).
Given:
Width of door (w) = 1.05 m
Force (F) = 23.0 N
The radius (r) can be calculated as half of the width of the door:
r = w / 2 = 1.05 m / 2 = 0.525 m
Plugging the values into the formula:
τ = 0.525 m * 23.0 N * sin(90 degrees)
τ = 0.525 m * 23.0 N * 1
τ = 12.075 Nm
Therefore, the torque resulting from applying a tangential force of 23.0 N to the outer edge of the door is 12.075 Nm.
2. In the second scenario, where the force is applied at a 45-degree angle:
The torque (τ) can be calculated using the same formula mentioned above:
τ = r * F * sin(θ)
Given:
Angle (θ) = 45 degrees
Using the same radius value as calculated before (r = 0.525 m), and plugging in the values into the formula:
τ = 0.525 m * 23.0 N * sin(45 degrees)
τ = 0.525 m * 23.0 N * √(2)/2
τ = 0.1838 Nm
Therefore, the torque resulting from applying a force of 23.0 N at a 45-degree angle to the outer edge of the door is 0.1838 Nm.