The discussion revolves around a physics problem related to mass, force, and acceleration, specifically focusing on the dynamics of a boy sitting on a chair. Participants are attempting to solve part 2 of the homework, which involves analyzing forces acting on the boy and the chair using free body diagrams and Newton's laws.
Some participants have provided guidance on the importance of considering all forces acting on the boy and chair, while others express uncertainty about their calculations and the application of Newton's laws. Multiple interpretations of the problem are being explored, with no explicit consensus reached.
There is mention of potential confusion regarding the roles of tension and normal forces in the system, as well as the need to clarify the mass contributions from the boy and chair in the calculations. Some participants question the appropriateness of their methods and results, indicating a lack of complete information or understanding.
very good!rodjav305 said:
Now you need to draw a free body diagram of the chair, in order to calculate the normal force of the boy on the chair, OR, draw a free body diagram of the boy, in order to determine the normal force of the chair on the boy. You can do either FBD, whichever one you find easier to identify the forces acting, then use Newton's laws. Note that the acceleration of the boy and the acceleration of the chair are the same as the acceleration of the system.
I am not sure how you answered part a so nicely, while for this part b you are for some reason subtracting masses in units of kg and ending up with a force in units of Newtons, and then ignoring Newton's laws as wellrodjav305 said:
Yes, it is wrong, I don't know what you are doing here.rodjav305 said:
. No.rodjav305 said:
No, incorrect. I have absolutely no idea how you solved part a unless it was given you. But as for part b, when you isolate a mass, like the chair, you examine all the forces acting on it, and use that mass alone in your net force =ma equation. I've already given you some hints in post #5.
No, that's the net force acting on the boy and chair that you correctly used in part a.rodjav305 said:
There's only one rope in the horse pull problem. In this problem, there are two rope forces acting on the child-swing system when you draw a FBD (Free Body Diagram) that isolates the child and swing from the pulley. This is not the case when you draw the FBD showing the forces on the child alone, since only one rope acts on him (at his hands).I Like Pi said:
If you are looking at the forces on the chair, then FT -Fgchair - N =mchair*a. You forgot the normal force of the boy acting down on the chair. If you are looking at the forces on the boy, then FT -Fgchild + N = mchild*a. Again, you forgot the normal force of the chair acting up on the child. FT is given as 213.5 N.I Like Pi said:
PhanthomJay said:
that makes sense, but what i don't get is that, i know for the boy himself, he has a normal force because he lies on the swing, which is the swing acting on the boy. What i don't get is why does the chair have a normal force? As it has nothing underneath it? Wouldn't the force of the boy be, the force of the boy? not the normal force? cause then its a reaction force By Newton's 3rd law, if the chair exerts a normal force on the boy, the boy must exert an equal and opposite normal force on the chair.I Like Pi said:Oh, thank you
But its got the kid on top of it! Normal forces act perpendicular to the surfaces in contact, and always push toward the object on which they act, if not physicaly connected to it.
The force of the boy is the normal force on the chair, acting down on it. The force of the boy on the chair is not his weight.
I guess it's an action/reaction force: the boy exerts an action force (normal force) on the chair, and the chair exerts an equal and opposite reaction force (normal force) on the boy.cause then its a reaction force
Thanks for your time!