Minimum force required to keep two blocks from not falling

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SUMMARY

The discussion focuses on the minimum force required to prevent two blocks from falling, specifically analyzing the frictional forces at play. The maximum friction force between the blocks is calculated as 80N (0.1 * 800N), while only 20N is necessary for support, confirming stability. The heavier block Q requires the maximum force, but calculations for block P are also deemed beneficial for clarity. This analysis highlights the importance of understanding friction in static equilibrium scenarios.

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nafisanazlee
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Homework Statement
Two blocks P and Q are of weight 20N and 100N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of static friction between the blocks is 0.1 and between block Q and the wall is 0.15, what will be the minimum force to keep the blocks in equilibrium?
I've tried to solve it in this way, but I'm not sure if my approach is correct or not. Can you please check?
Relevant Equations
Fsmax = μsN
CamScanner 11-26-2023 02.56.jpg
 
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:welcome:

Looks right. You might want to add why block P does not slide.
 
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PeroK said:
:welcome:

Looks right. You might want to add why block P does not slide.
because the maximum friction force that can be provided between the two blocks becomes 0.1*800= 80N, and we only need 20N for support, so it's fine..?
 
nafisanazlee said:
because the maximum friction force that can be provided between the two blocks becomes 0.1*800= 80N, and we only need 20N for support, so it's fine..?
Yes, it was fairly obvious from the numbers that the maximum force was needed for Q (as it is much heavier). But, it does no harm to show the calculation for P as well.
 
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PeroK said:
Yes, it was fairly obvious from the numbers that the maximum force was needed for Q (as it is much heavier). But, it does no harm to show the calculation for P as well.
Thank you so much for your time. Much appreciated.
 
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