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

In summary, three blocks with masses of 5, 10, and 25 kg are arranged on a frictionless surface with a rightward force of 14 N applied to the left-most block. The magnitude of the force that the middle block exerts on the rightmost one is 8.75 N, and the magnitude of the force that the leftmost block exerts on the middle one is 2.8 N. When the left-right order of the blocks is reversed, the magnitude of the force that the leftmost block exerts on the middle one is still 2.8 N.
  • #1
farrah003
15
0

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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  • #2
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:
 
  • #3
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
 
  • #4
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
 
  • #5
I'm confused :confused: …

show your equation for part b. :smile:
 
  • #6
F = (5)(.35) = .175
 
  • #7
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.
 
  • #8
so then what formula do i use ?
 
  • #9
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.
 

1. What is the formula used in this problem?

The formula used in this problem is Fm1m2 = Fm3m2 - 14Force on Blocks, where Fm1 and Fm2 represent the forces acting on the first block, Fm3 represents the force acting on the second block, and 14Force on Blocks is a constant value.

2. How do you find the magnitudes and directions of the forces?

To find the magnitudes and directions of the forces, you will need to use vector addition. Use the given formula and plug in the values for Fm1, Fm2, and Fm3 to find the resultant force, which will give you the magnitude and direction of the force acting on the blocks.

3. What do the subscripts in the formula represent?

The subscripts in the formula represent the different blocks involved in the problem. Fm1 and Fm2 represent the forces acting on the first block, while Fm3 represents the force acting on the second block.

4. How does the constant value affect the problem?

The constant value, 14Force on Blocks, represents a force that is constant and acting on both blocks. This force can be either in the same direction or in the opposite direction of the other forces, depending on the specific problem. Including this constant value allows for a more accurate calculation of the resultant force.

5. Can this formula be used for any force problem involving blocks?

Yes, this formula can be used for any force problem involving blocks as long as the forces are acting on different blocks and there is a constant force acting on both blocks. However, the specific values for Fm1, Fm2, and Fm3 will vary depending on the given problem.

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