What Is the Minimum Force Needed to Keep Two Blocks Stationary?

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SUMMARY

The discussion centers on calculating the minimum force required to keep two blocks stationary under different conditions of friction. For part (a), with no friction between the second block and the floor, the minimum force is derived as ##F=mg\frac{m+M}{uM}##, where ##m=1[kg]##, ##M=5[kg]##, and ##u=0.6##. In part (b), where friction is present, the relationship between the applied force and the frictional force must be carefully considered, leading to the equation ##a=\frac{F-u_pMg}{m+M}##. The confusion arises from the assumption that the force between the blocks is equal to the force applied to the first block, which is incorrect.

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doktorwho
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Homework Statement


Diagram.JPG

##m=1[kg]##, ##M=5[kg]##, ##u=0.6##
Calculate the minimum force required for the blocks to remain in the same position thourought the motion. For the part a) there is no friction between block 2 and the floor, for the part b there is and its ##u_f##

Homework Equations


3. The Attempt at a Solution [/B]
For the part a) the frictional force between the blocks must be equal to ##mg##. The frictional force is proportional to the force on the block.
##mg=F_{fr}##
##mg=uN## ##\Rightarrow## ##N=ma## ##\Rightarrow## ##F=(m+M)a##
##mg=um\frac{F}{m+M}## ##\Rightarrow## ##F=g\frac{m+M}{u}##
The book solution is: ##F=mg\frac{m+M}{uM}##
I was certain i got this right and then i see a different answer in the book. What is wrong?
For the part b)
##mg=F_{fr}##
##mg=uN## ##\Rightarrow## ##N=ma## ##\Rightarrow## ##(m+M)a=F-u_pMg## ##\Rightarrow## ##a=\frac{F-u_pMg}{m+M}##
Is this right so far?
 
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doktorwho said:
I was certain i got this right and then i see a different answer in the book. What is wrong?
You are assuming that the force between the blocks equals the force applied to the first block. Not so.
 
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