NLM Problem: More variables than equations

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The discussion revolves around solving a problem involving two blocks, where the equations derived from free body diagrams (FBDs) result in more variables than equations. The participants highlight that the initial FBD is incorrect, particularly regarding the treatment of accelerations as forces. They emphasize the need to consider the system as a whole to determine the net horizontal force and acceleration before analyzing each block separately. The conversation suggests that accurate labeling of coordinates and careful consideration of forces is crucial for finding a valid solution. Ultimately, participants are encouraged to post their revised solutions for clarity and future reference.
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Homework Statement
Two blocks initially at rest are kept in contact as shown in the figure. A force F is applied on the left block. The blocks are of masses ##m_1## and ##m_2##. Find the acceleration of the blocks. All surfaces are smooth.
Relevant Equations
##Force = Mass \times Acceleration##
fbd1.png

Attempt at Solution::

##m_2## is chosen as reference frame and FBD is drawn as shown above. We get the following equations:

From ##m_1## by choosing axes along and perpendicular to acceleration of ##m_1## w.r.t ##m_2## (##a'##):

$$
m_1 a' + m_1 a \cos{\theta} + m_1 g \sin{\theta} = N_1 \sin{\theta} + F \cos{\theta}
$$
$$
N+N_1 \cos{\theta} + m_1 a \sin{\theta} = m_1 g \cos{\theta} + F \sin{\theta}
$$

From ##m_2## by choosing horizontal and vertical axes:
$$
N_2 = N \cos{\theta} + m_2 g
$$
$$
m_2 a = N \sin{\theta}
$$

Which gives us four equations, but five variables, namely ##a',a,N_1 ,N_2, N##
 
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Your FBD is incorrect. ##m_1a## and ##m_2a## are not forces. If you consider the two blocks together as your system, what is the net horizontal force acting on that system? What is its horizontal acceleration? Start from there then draw two separate FBDs for each block.
 
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Please label the coordinates on the diagram, and the associated coordinate with each equation you have written down.
 
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kuruman said:
Your FBD is incorrect. ##m_1a## and ##m_2a## are not forces. If you consider the two blocks together as your system, what is the net horizontal force acting on that system? What is its horizontal acceleration? Start from there then draw two separate FBDs for each block.
umm.. ##m_1a## and ##m_2a## are pseudo forces because accelerating frame ##m_2## (with acceleration a towards right) is my Reference frame.

Net horizontal force on the two blocks as a system is F.
Sum of mass × horizontal acceleration of two blocks is F.

But how can you comment about their individual horizontal acceleration?
 
Think about a' and N1. can both be nonzero?
 
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Why ##N_1## participating in the sum of the forces if you assume the small mass is accelerating up the incline?

There are probably a few errors here. You should do as @kuruman suggest and present accurate FBD's of each block.
 
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haruspex said:
Think about a' and N1. can both be nonzero?
I think this actually solves it. Ok trying and editing to final solution...
 
So what did you get for ##a## and ##a'## as labeled in your diagram?
 
Seeker220 said:
I think this actually solves it. Ok trying and editing to final solution...
Please post the edited solution in a new post in this thread.
That's preferrable to editing Post #1 itself, which gets very confusing for anybody reading the thread at a later time.
 
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