Is This the Solution? A Review of Dynamics Pushing Blocks with Visuals

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Homework Help Overview

The discussion revolves around a dynamics problem involving two blocks, where participants analyze free body diagrams (FBDs) and equations of motion. The original poster seeks validation for their calculations related to forces and accelerations acting on the blocks.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants present equations of motion for two masses, discussing the relationships between normal forces, friction, and applied forces. Some participants express confusion regarding specific equations and seek clarification on missing components.

Discussion Status

There is ongoing dialogue about the accuracy of the equations and calculations presented. Some participants have offered corrections and suggestions for improvement, while others are working to clarify their understanding of the problem setup. Multiple interpretations of the equations are being explored.

Contextual Notes

Participants note that the textbook does not provide answers, which adds to the uncertainty in verifying their solutions. There is also mention of potential typographical errors in the equations that could affect the results.

alingy1
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Please look at pictures.
Weirdly, my textbook does not give the answer to this question.

Can you check my answer? Sorry for the really bad writing. If you ask, I can more clearly write it.
 

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The handwriting is decipherable, but it's hard to follow what you are doing because it's not clear which equations relate to which FBD.
Please post your working by typing - it's not hard - making it clear which forces relate to which body.
Click on the Go Advanced button. You can pick ∑ off the quick symbols on the right and superscript and subscript from the X2, X2 icons along the top.
 
Here goes nothing!
For m1:
∑Fy=N1-m1g-sin(30)F=0
∑Fx=-N21-fk+cos(30)F=m1a

For m2:
∑Fx=-μkm2g+N12=m2a
∑Fy=N2-m2g=0

Add both ∑Fx equations:
km1g-μkm2g+cos(30)F=(m1+m2)a

a=1.41 m/s^2
N12=N21=11.52N

That looks great! How about the physics?
 
You've missed something out. Write out the equations relating N1, F, fk.
 
I do not understand. Can you be more specific? What is it I am missing? I've been looking at the problem long enough that everything seems good.
 
Got it. I mixed up the fk equation. I'm fixing it now and posting the new solution.
 
alingy1 said:
Here goes nothing!
For m1:
∑Fy=N1-m1g-sin(30)F=0
∑Fx=-N21-fk+cos(30)F=m1a

For m2:
∑Fx=-μkm2g+N12=m2a
∑Fy=N2-m2g=0

Add both ∑Fx equations:
km1g-μkm2g+cos(30)F=(m1+m2)a

a=1.41 m/s^2
N12=N21=11.52N

That looks great! How about the physics?


Here is my new equation:

μk(sin(30)F+m1g)-μkm2g+cos30F=(m1+m2)a

a=1.16 m/s^2
N12=N21=10.52 N
 
alingy1 said:
Here is my new equation:

μk(sin(30)F+m1g)-μkm2g+cos30F=(m1+m2)a

a=1.16 m/s^2
N12=N21=10.52 N

Those answers are correct (and if they're not, we made the same mistakes hahah)

Except for the rounding, if you rounded a to 2 decimals it would actually go up to 1.17 but that's pedantic, your equation is correct.



edit: Actually, there should be a negative sign in very the beginning of your first equation, but you must have just forgotten to type that, because your answer is correct.
 
Yes, I did forget the minus! I hope someone else confirms! At least we are two ;)
 
  • #10
alingy1 said:
I hope someone else confirms! At least we are two ;)
Make that three.
 
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