The discussion revolves around a problem related to the forces acting on slip-joint pliers, specifically focusing on the free body diagram (FBD) and the internal forces at play. Participants are examining the implications of a 30 lb force and how it interacts with the forces on the pin and the circular component of the pliers.
The conversation is ongoing, with participants exploring different interpretations of the forces involved. Some guidance has been offered regarding the necessity of considering internal forces when analyzing the FBD, but there is no explicit consensus on the correct representation of forces.
Participants are working within the constraints of a specific homework problem, which may limit the information available for discussion. The nature of the problem requires careful consideration of forces and their directions, leading to questions about assumptions made in the FBD.
since the force on the pin is internal, you have to 'take apart' the pliers in a 'free body diagram' of one of the 2 pieces, in order to solve for the internal forces. When you do this, the other 30 lb force is not in the picture. The force of the circle thing on that part of the pliers is perpendicular (normal) to the contact point, and can be broken into its horizontal and vertical components.doubled said:
Well yes, that's Newton's third law, the force of the pliers on the circle act to squeeze it . But when you draw the FBD of the pliers, the force of the circle thingy acting on the pliers points in the +x and -y direction (down and to the right), which you can show by summing moments about the pin.doubled said:
Well, you can't just press down on the upper part, you have to sqeeze the handles together with an equal but opposite force on each in order to make the thing work. The lower jaw moves up, and the upper jaw moves down, as if it were biting the 'circle' (a nut or bolt or rod, etc).
