Force pushes three blocks, find mag of force exerted on last two blocks

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Homework Help Overview

The problem involves three blocks in contact being pushed across a rough horizontal surface by a force of 75 N. The objective is to find the magnitude of the force exerted on the 3.5 kg block by the 4.1 kg block, considering the effects of friction with a coefficient of 0.076.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss treating the three blocks as a single system to analyze the forces acting on them. There are attempts to apply Newton's laws and consider the effects of friction in the calculations. Questions arise regarding the correct application of forces and the direction of friction.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces involved. Some have provided hints and guidance on how to approach the problem, while others are questioning their understanding of the setup and calculations.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also confusion regarding the labeling of the blocks and the direction of forces, which is affecting their reasoning.

gap0063
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Homework Statement


Three blocks in contact with each other are
pushed across a rough horizontal surface by a
75 N force as shown.
The acceleration of gravity is 9.8 m/s2 .
[PLAIN]http://img178.imageshack.us/img178/6198/007ow.png

If the coefficient of kinetic friction between
each of the blocks and the surface is 0.076,
find the magnitude of the force exerted on the
3.5 kg block by the 4.1 kg block.
Answer in units of N.

Homework Equations


sum of Fy=N-mg=0 => N=mg
sum of Fx= -fk=max
fk=uN


The Attempt at a Solution


I have no clue how to figure out the force just between those two block...

Do I add the first two blocks together to act a one force against the third block?
 
Last edited by a moderator:
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Hint: Start by treating the three blocks as a single system. What can you deduce?
 
Doc Al said:
Hint: Start by treating the three blocks as a single system. What can you deduce?

You mean like m1+m2+m3=9.6 kg

Then the forces acting in the y direction is N1+N2+N3 and (m1+m2+m3)g
forces acting in the x direction is fk and F

which fk=u*(N1+N2+N3)

right?
 
gap0063 said:
You mean like m1+m2+m3=9.6 kg

Then the forces acting in the y direction is N1+N2+N3 and (m1+m2+m3)g
forces acting in the x direction is fk and F

which fk=u*(N1+N2+N3)

right?
Right. Keep going. Apply Newton's 2nd law.
 
Doc Al said:
Right. Keep going. Apply Newton's 2nd law.

so the sum of Fy= N1+N2+N3-Mg=0, where M is m1+m2+m3
so N1+N2+N3=Mg=> where N1+N2+N3=94.08

so the sum of Fx= -fk=Ma

fk=u(N1+N2+N3)
fk=7.15008

so -7.15008=9.6a
a=-0.7448 m/s^2
this is small... is this right?
 
gap0063 said:
so the sum of Fy= N1+N2+N3-Mg=0, where M is m1+m2+m3
so N1+N2+N3=Mg=> where N1+N2+N3=94.08
OK.

so the sum of Fx= -fk=Ma
ΣF = ma
But ΣF ≠ -fk. Friction is not the only horizontal force acting.

fk=u(N1+N2+N3)
fk=7.15008
OK.

so -7.15008=9.6a
a=-0.7448 m/s^2
this is small... is this right?
No, as explained above.
 
Doc Al said:
OK.


ΣF = ma
But ΣF ≠ -fk. Friction is not the only horizontal force acting.


OK.


No, as explained above.

Okay so where does fk play into?
 
Last edited:
gap0063 said:
Okay so where does fk play into?
Friction is one of the horizontal force acting on the blocks. What direction does it act?

What other horizontal forces act? Look at your diagram.
 
Doc Al said:
Friction is one of the horizontal force acting on the blocks. What direction does it act?

What other horizontal forces act? Look at your diagram.

is it sum of Fx= fk+F=ma?
 
  • #10
gap0063 said:
is it sum of Fx= fk+F=ma?
Yes, but make sure you have the correction directions and thus the correct signs for the forces. (Hint: Let 'right' be positive and 'left' be negative.)
 
  • #11
Doc Al said:
Yes, but make sure you have the correction directions and thus the correct signs for the forces. (Hint: Let 'right' be positive and 'left' be negative.)

Okay, so I did -fk+F=Ma

-7.15008+75=9.6a => a=7.0677 m/s^2 (wrong)

what did I do wrong?
 
  • #12
gap0063 said:
Okay, so I did -fk+F=Ma

-7.15008+75=9.6a => a=7.0677 m/s^2 (wrong)

what did I do wrong?
Looks fine to me. Why do you say it's wrong? Did the problem even ask you for the acceleration? (Finding the acceleration is just a step towards the required answer.)
 
  • #13
Doc Al said:
Looks fine to me. Why do you say it's wrong? Did the problem even ask you for the acceleration? (Finding the acceleration is just a step towards the required answer.)

You're right I need the force between m2 and m3...

So can I do Sum of F= (m2+m3)*a?
 
  • #14
gap0063 said:
So can I do Sum of F= (m2+m3)*a?
That won't tell you anything about the forces between m2 and m3. Hint: Analyze m3 alone or m1 + m2 together.
 
  • #15
Doc Al said:
That won't tell you anything about the forces between m2 and m3. Hint: Analyze m3 alone or m1 + m2 together.

So the only forces acting on m3 in the x direction is the force from m1 and m2...
in the y direction it is N and m3g

So for the force in the x direction from m1 and m2 (i'll call P12)

F-P12=m3a?
I'm not sure how to calculate P12...
 
  • #16
gap0063 said:
So the only forces acting on m3 in the x direction is the force from m1 and m2...
No. What about friction? Which block is m3? Your diagram isn't labeled.
in the y direction it is N and m3g
OK.

So for the force in the x direction from m1 and m2 (i'll call P12)

F-P12=m3a?
I'm not sure how to calculate P12...
Once you account for all the forces, the only unknown will be the force between m2 and m3. You'll solve for it.
 
  • #17
Doc Al said:
No. What about friction? Which block is m3? Your diagram isn't labeled.

OK.


Once you account for all the forces, the only unknown will be the force between m2 and m3. You'll solve for it.

So then sum of F = F-P12-fk=ma?

and P12 is my force bwtween m2 and m3 right?
 
  • #18
gap0063 said:
So then sum of F = F-P12-fk=ma?

and P12 is my force bwtween m2 and m3 right?
Right. (Assuming m3 is the leftmost block.)
 
  • #19
Doc Al said:
Right. (Assuming m3 is the leftmost block.)

Sorry m3 is the right block I'm trying to figure out the right block and middle


so if it is the right block does force even apply?

so I did the sum of Fx = -P12-fk=m3a is that right?
 
  • #20
gap0063 said:
Sorry m3 is the right block I'm trying to figure out the right block and middle
OK, that's different.

so if it is the right block does force even apply?
The applied force F acts on m1, not on m3.

so I did the sum of Fx = -P12-fk=m3a is that right?
Which direction does m2 push on m3?
 

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