Block Motion and Normal Force

  • Thread starter Nicolaus
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  • #1
Nicolaus
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Homework Statement


You are pushing four blocks of masses 4, 7, 11, and 17kg positioned next to each other on a frictionless surface.
Find the force of contact between the 7 and 11kg blocks when you push the blocks with a 190N force to the left.
They are arranged as such: |17||11||7||4|


Homework Equations


F = 190N = (17+11+7+4)a
a = 4.87

The Attempt at a Solution


m4a = F - F7on4
F7on4 = 190 - (4x4.87) = 170.52N
F4on7 = -170.52N
m7a = F4on7 - F11on7
F11on7 = -170.52 - (7x4.87) = -204.12N
 

Answers and Replies

  • #2
Doc Al
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You're making a sign error. Note that acceleration is to the left.
 
  • #3
Nicolaus
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So F7on4 should be -170.52 given that I have chosen the left direction as positive?
 
  • #4
Doc Al
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So F7on4 should be -170.52 given that I have chosen the left direction as positive?
Yes. F7on4 points to the right, which is negative in your convention.
 

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