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

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Homework Help Overview

The problem involves three blocks of mass 5 kg, 10 kg, and 25 kg on a frictionless surface, with a force of 14 N applied to the leftmost block. Participants are tasked with finding the magnitudes of forces exerted between the blocks under different conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using Newton's second law (F = ma) to find the acceleration of the system and the forces between the blocks. There are attempts to calculate the total mass and resulting acceleration, but confusion arises regarding the application of forces on individual blocks.

Discussion Status

Some participants have made attempts to calculate the acceleration and forces, but there is uncertainty about the correct application of the equations. Guidance has been offered regarding the forces acting on the blocks and the need to consider both applied and reaction forces.

Contextual Notes

Participants express confusion about the equations and the relationships between the forces acting on the blocks, indicating a need for clarification on how to approach the problem systematically.

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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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