A rotational axis is directed perpendicular to the plane of...

A rotational axis is directed perpendicular to the plane of a square and is located as shown in the drawing. Two forces, $\overrightarrow{\mathbf{F}}_{1}$ and $\overrightarrow{\mathbf{F}}_{2},$ are applied to diagonally opposite corners, and act along the sides of the square, first as shown in part a and then as shown in part b of the drawing. In each case the net torque produced by the forces is zero. The square is one meter on a side, and the magnitude of $\overrightarrow{\mathbf{F}}_{2}$ is three times that of $\overrightarrow{\mathbf{F}}_{1} .$ Find the distances a and b that locate the axis.

Key Concepts

Torque

Torque is a measure of the tendency of a force to produce rotation about a particular axis. It is defined as the cross product of the position vector (from the axis to the point of force application) and the force vector, capturing both the magnitude of the force and the lever-arm distance, as well as the angle between the force and the position vector.

Lever Arm

The lever arm is the perpendicular distance from the axis of rotation to the line of action of a force. It is a critical factor in determining the magnitude of the torque produced by a given force, as only the component of the force acting perpendicular to the lever arm contributes to rotation.

Rotational Equilibrium

Rotational equilibrium occurs when the net torque acting on a system is zero, meaning there is no resultant angular acceleration. This condition ensures that the object either remains at rest or rotates with constant angular velocity, which is a common scenario in statics problems.

Force Couple

A force couple consists of two forces that are equal in magnitude, opposite in direction, and separated by a distance, which create a pure torque without resulting in any net force. This concept is important in understanding how forces can balance each other to produce a state of zero net torque while still generating rotational effects.

Transcript

00:01 Okay, so we have this system here.

00:04 Now in picture a, we have that force 1 acts on the square at distance b here.

00:14 So since force 1 is perpendicular to this length b here, then the torque due to the torque due to force 1 is just going to be force 1 times b.

00:26 So f1 times b.

00:29 Now torque 2 here, it is perpendicular to this length a here.

00:35 So the torque is just going to be f2 times a.

00:39 And remember f2 is three times the magnitude of f1.

00:42 So we have that force 2 equals 3 f1.

00:45 Or the torque due to force 2 is going to be 3 f1a.

00:50 So since the net torque is equal to 0, we set the two torques equal to each other.

00:55 So we have f1b is equal to to 3f1a.

01:00 The f1s can cancel out on both sides.

01:04 So we have here that f or that b is equal to 3a.

01:11 Now here on the picture b now we have that force one is acting on this distance.

01:20 Here this is going to be length.

01:22 All right.

01:24 This here is going to be length 1 minus a since the total distance of the square is 1 meter.

01:32 Right.

01:32 So now we have that torque 1 is going to be equal to the force 1 times 1 minus a...

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