Solving Horizontal Force Problems Involving Blocks

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Homework Help Overview

The discussion revolves around a physics problem involving two blocks subjected to a horizontal force. The participants explore the relationship between the forces acting on the blocks and their resulting acceleration, with specific values provided for the forces in different scenarios.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the net force acting on the blocks and how to relate it to their acceleration. They attempt to set up equations based on the forces and masses involved, questioning how to express the forces in terms of acceleration.

Discussion Status

There is ongoing exploration of the relationships between the forces and the masses of the blocks. Some participants have provided equations and attempted to derive values for acceleration and force, while others are seeking clarification on how to calculate the applied force.

Contextual Notes

Participants are working under the constraints of the problem statement, which specifies the forces acting on the blocks and their combined mass. There is a focus on ensuring that the acceleration is consistent across both blocks.

jarny
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Homework Statement



In Figure 5-50a, a constant horizontal force F->a is applied to block A, which pushes against block B with a 23.0 N force directed horizontally to the right. In Figure 5-48b, the same force is applied to block B; now block A pushes on block B with a 11.0 N force directed horizontally to the left. The blocks have a combined mass of 12.0 kg. What are the magnitudes of (a) their acceleration in Figure 5-50a and (b) force F->a?


Homework Equations



Fnet=m*a

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c05/fig05_50.gif


The Attempt at a Solution



Ok so for block scenario one I have to find the Fnet to solve for m*a.
So I have Fnet= sum of forces=m*a

for case two

Fnet= sum of forces =m*a


So I have

12*a= Fapp+23-F_AB
12*a=Fapp-11+F_AB


I am stuck right here any tips?
 
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Write down what you know.

Ma + Mb = 12

23N = Mb*a

11N = Ma*a

Since a is the same, then

Mb/Ma = 23/11

Mb = 12 - Ma = 23/11*Ma

Ma = 12*11/34 ...
 
Thank very much
 
Wait how would I found F->


Do I use: 23=F-> + X and 11=F-> -X ?


Then F->= 17 N?
 
What is your total mass and the acceleration of the system?
 
So the F->applied would be 12*a which I found?
 
jarny said:
So the F->applied would be 12*a which I found?

Isn't a the acceleration of both blocks?

And total mass is Ma + Mb.

So ...
 
Thanks
 

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