Understanding Net Force and Normal Force in a Frictional System

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Homework Help Overview

The discussion revolves around understanding the concepts of net force and normal force in a system involving friction between two blocks. Participants are exploring the relationships between forces acting on the blocks and how to calculate the normal forces involved.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of net force and the role of normal force in the context of friction. Questions arise regarding the determination of acceleration and the necessity of considering both blocks in the system.

Discussion Status

Some participants have offered insights into calculating normal forces and balancing forces acting on the blocks. There is an ongoing exploration of the problem, with multiple interpretations of how to approach the calculations and the relationships between the forces.

Contextual Notes

There is mention of potential confusion regarding the normal forces acting on both blocks and the implications of friction in the calculations. Participants are also considering the effects of weight and the coefficient of friction in their reasoning.

muna580

Homework Statement



http://img255.imageshack.us/img255/9655/untitled1vq1.jpg

The Attempt at a Solution



Okay, so I have no idea how to do this problem. Well first of all, the net force of the whole system is F - f = (m1 + m2)a. I have no idea what a is. Also, since friction is involved we need to find the normal force. How do I find the normal force?
 
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wow that's a wild and wooly problem. first, word normal here is in the perpendicular sense. so when two bodies are in contact, its the reactive force (Newtons third law) in response to our weight on the floor which keeps us from falling thru it. The floor pushes back. Curiously common sense fails us here as frictional forces have nothing to do with surface area, only the magnitude of the force keeping them in contact times that coefficient, mu.

So here we have two normal forces at right angles.

Lets start with the easy half, the normal force between the big block and its supporting surface. Thats easy, Fn=mg*mu=9.8*4.2*0.24= 9.87 (weight times friction)

Now the other half of the problem is less intuitive. But in principle no different, the opposite of the force of the big block on small block.

(Here I am a bit fuzzy, do we have to worry about both blocks or just the big one.) No worries, If wrong someone will come to our rescue.

Lets take the case where we only concern ourselves with the big block;

now we know that in order for the smaller block not to fall, the force impressed on it must be equal to .7kg*9.8*.38=2.61N

the Sum of forces=ma must then be 2.61

So Fx-Fn=2.61 where Fn=9.87

Adding the two gives 12.49. May be dead wrong, but hope its of some help. Seems like a very complex problem for someone with no idea how to compute Normal force.:rolleyes:
 
Oh I get it now. Thanks a lot. I have to slove for normal forces on both blocks. Thanks a lot.
 
NP. let me know if the answer was close. First time I seen one like that.
 
If you look at a FBD of the small block, the friction force between the 2 blocks must be balanced by the small block's weight, and therefore equal to 0.7g Newtons. You should now be able to calculate the Normal force (N) between the blocks in the positive x direction, and then use Newton 2 to solve for the acceleration.
 

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