The discussion revolves around a pulley-block-spring system, specifically focusing on the forces acting on a hanging block. Participants are examining the dynamics involved, particularly the roles of tension, spring force, and gravitational force in the context of a free body diagram (FBD).
The discussion is ongoing, with participants providing guidance on the importance of free body diagrams and questioning assumptions about the forces involved. There is a mix of interpretations regarding the setup of the problem and the forces acting on the blocks.
Some participants note that the original poster's reference to a specific moment in a video may lead to confusion, suggesting that clearer problem statements and diagrams would aid in understanding. There is also mention of the need to consider the direction of forces in the equations presented.
No. Think what it should be when the wire is attached to a fixed point.Abhimessi10 said:In FBD of 2nd block it should have been T+kx-mg=ma right?
Sorry but i don't understand what you meanBvU said:Un-PF format and ambiguous:
"why doesn't the block have a spring force "there are two blocks. The spring is not attached to the left block.
No. Think what it should be when the wire is attached to a fixed point.
Chandra Prayaga said:Draw the free body diagram of each block. What are the forces on the 15 kg block? What is the net force? Then use Newton's second law to calculate the unknown force. Now mov to the second block and do the same thin.
At 5:32 for me if the second block attached to spring is going down,then spring force and tension should act on the blockChandra Prayaga said:It is still a problem of not drawing a proper force diagram. In the equation which you gave, what is T? What is kx? What is mg, what is its direction? All these will become clear if you draw the FBD.
The question I'm talking about is at 5:32 which involves a pulleyChandra Prayaga said:1. Why is the second block going down? The diagram shows it on the table.
2. The tension is coming from which object?
The wire exerts tension ,spring exerts Kx in same direction so FBD for block 2 will be mg-(T+kx)=ma in my opinionChandra Prayaga said:Sorry. I did not realize that you were talking about a different diagram later in the video. OK.
You are writing two separate forces, T and kx. What is exerting these two forces?
But if the block is going down what is causing the restoring force kx of springChandra Prayaga said:The wire and the spring are not two separate things, and each one is not separately attached to the block. The tension on the block is because of the spring, so T is the same as kx. Both should not appear separately in your equation.