In the free body diagram of the board and weight system, you look at external forces only , the weight forces and pivot support forces. The normal forces of the weights are internal forces and excluded. If you wish , however, you can draw a free body of the board alone , then the forces external to the board become the pivot force and the normal force of the 2 objects acting on the board, and get the same result.Gabe805 said:I was just curious as why, when drawing a free-body diagram of a board about a pivot with weight on either side of the pivot, we do not consider the corresponding normal forces to those weights(using torque considerations). This question was already asked in another post but was asked a little less clearly.
Torque is a measure of the force that causes an object to rotate around an axis. It is the product of the force applied and the distance from the axis of rotation.
The normal force is the perpendicular force exerted by a surface on an object in contact with it. In the context of torque, the normal force is the force that acts as the pivot point for the object to rotate. Without a normal force, there would be no torque.
The magnitude of torque is directly proportional to the magnitude of the normal force. This means that the greater the normal force, the greater the torque, and vice versa.
Yes, torque can be negative. This occurs when the force applied and the distance from the axis of rotation are in opposite directions, causing a rotational motion in the opposite direction.
In a free-body diagram, torque is represented by a curved arrow, with the direction of the arrow indicating the direction of the rotational motion. The length of the arrow represents the magnitude of the torque, with a longer arrow indicating a greater torque.