What is the magnitude of the normal force in this block stacking scenario?

AI Thread Summary
In a block stacking scenario on a frictionless surface, three blocks with masses of 37.0 kg, 18.0 kg, and 20.0 kg are analyzed under a downward force of 170 N applied to the top block. The normal force exerted by the bottom block on the middle block is calculated using the equation Fn + T - mg = 0. Initially, the total weight of the two upper blocks is calculated as 372.4 N. After correcting for the downward force, the normal force is determined to be 542.4 N. This calculation highlights the importance of accurately accounting for all forces acting on the blocks.
lacar213
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Homework Statement


Three blocks are arranged in a stack on a frictionless horizontal surface. The bottom block has a mass of 37.0 kg. A block of mass 18.0 kg sits on top of it and a 20.0 kg block sits on top of the middle block. A downward vertical force of 170 N is applied to the top block. What is the magnitude of the normal force exerted by the bottom block on the middle block?



Homework Equations


Fn + T + (-mg) = 0


The Attempt at a Solution


20 + 18 = 38*9.8 = 372.4 N
Fn + 170 - 372.4 = 0
Fn = 202.4
 
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lacar213 said:

The Attempt at a Solution


20 + 18 = 38*9.8 = 372.4 N
Good.
Fn + 170 - 372.4 = 0
Careful: That 170 N force acts downward.
 
So the 170 needs to be negative .. ?
Fn + (-170) - 372.4 = 0
Fn = 542.4
 
Looks good!
 

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