Fm1m2 = Fm3m2 - 14Force on Blocks: Find Magnitudes & Directions

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The discussion focuses on a physics problem involving three blocks of different masses on a frictionless surface, with a 14 N force applied to the leftmost block. Participants are trying to determine the forces exerted between the blocks, specifically the force the middle block exerts on the rightmost one and the force the leftmost block exerts on the middle one. There is confusion regarding the application of Newton's second law, particularly in calculating acceleration and the net forces acting on each block. Key insights include that all blocks share the same acceleration and that the left block experiences forces from both the applied force and the reaction force from the middle block. The conversation emphasizes the need to correctly apply the equations of motion to solve for the forces accurately.
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Homework Statement



Blocks of mass 5, 10, and 25 kg are lined up from left to right in that order on a frictionless surface so each block is touching the next one. A rightward-pointing force of magnitude 14 N is applied to the left-most block.


Homework Equations



[a] What is the magnitude of the force that the middle block exerts on the rightmost one?
What is the magnitude of the force that the leftmost block exerts on the middle one?
[c] Suppose now that the left-right order of the blocks is reversed. Now find the magnitude of the force that the leftmost block exerts on the middle one?

The Attempt at a Solution



this is what i tried doing but i think I am totally off
Fm3m2 = m2a + 14 - m1a
 

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Welcome to PF!

Hi farrah003! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)

They all have the same acceleration, so use F = ma on all-three-together to find a.

Then use F =ma (with that a) on individual blocks. :wink:
 
i tried f = ma
this is what i did
m = 5+10+25 = 40
14 = 40(a)
a = .35

but it said that's not the right answer
 
sorryy my mistake i got it for part a which was 8.75 but when i did it again for part b it said wrong
 
I'm confused :confused:

show your equation for part b. :smile:
 
F = (5)(.35) = .175
 
That F is the total force.

Apart from the right-most block (which you solved for in part a), each block has two forces on it.
 
so then what formula do i use ?
 
Ftotal = ma
 
  • #10
so then i do 14=5a
then the a is 2.8 n if i multiply that I am going to get 14 again ... I am confused !:confused:
 
  • #11
farrah003 said:
so then i do 14=5a
then the a is 2.8 n if i multiply that I am going to get 14 again ... I am confused !:confused:

No, a is the same for all three blocks.

The left block has two forces on it: 14 from the left, and the reaction force from the right.

The vector sum of those two forces is 5a.
 
  • #12
lol i still don't know what numbers to use in what formula
 
  • #13
(just got up :zzz: …)

The left block has two forces, 14 and R (the reaction force from the middle block), so F = ma means 14 - R = ma.
 
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