How Is the Contact Force Between the 7kg and 11kg Blocks Calculated?

[Nicolaus]

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

 

[Doc Al]

You're making a sign error. Note that acceleration is to the left.

 

[Nicolaus]

So F7on4 should be -170.52 given that I have chosen the left direction as positive?

 

[Doc Al]

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.