Minimum force required to keep two blocks from not falling

AI Thread Summary
The discussion centers on calculating the minimum force required to prevent two blocks from falling, focusing on the frictional forces involved. The maximum friction force between the blocks is determined to be 80N, while only 20N is needed for support, confirming stability. It is noted that the heavier block Q requires the maximum force, but calculating for block P is also deemed beneficial. The conversation emphasizes the importance of understanding the friction dynamics between the blocks. Overall, the calculations and reasoning provided clarify the conditions under which the blocks remain stable.
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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