Static friction and maximum/minimum force

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A block on an inclined plane with static friction is analyzed to determine the maximum horizontal force that can be applied without causing it to move. The discussion revolves around understanding the equations for maximum force, specifically focusing on the differences in the numerator involving static friction coefficients. Participants clarify their confusion regarding the terms in the equations, particularly the roles of sin and cos components. After reviewing the Free Body Diagram, the original poster resolves their confusion and expresses gratitude for the assistance. The thread highlights the importance of visual aids in understanding physics problems involving friction and inclined planes.
Poetria
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Homework Statement



A block of mass m is at rest on an inclined plane of angle θ with respect to the horizontal. There is friction between the block and the inclined surface (μk and μs).

Now, assume that the value of θ is less than the value in part A) (so that the block will remain stationary if no other forces act on it). What is the maximum magnitude of a horizontal Force F that(with respect to ground) can be applied without causing the block to move up the incline?

Homework Equations

The Attempt at a Solution



I have found a similar thread here:
https://www.physicsforums.com/threads/block-wedge-problem.698343/#post-4424430
and here: https://www.physicsforums.com/threa...-friction-min-and-max-forces-involved.619902/

I understand everything except one significant detail. I have to choose between several options:
mg(sinθ−μ*cosθ)/(cosθ+μ*sinθ)


or mg(sinθ+μ*cosθ)/(cosθ+μ*sinθ)

Friction is static. I don't understand the numerator. Why is there -/+μ*sinθ or -/+μ*cosθ?? Could you give me a hint?

 
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Why don't you draw a Free Body Diagram of the block on the incline
 
Oh, you are right. I did it. I understand it now. Silly me.
Many thanks. :)
Block and wedge.png
 
Poetria said:
Oh, you are right. I did it. I understand it now. Silly me.
Many thanks. :)

You did it yourself mate! Cheers!
 
Poetria said:
Oh, you are right. I did it. I understand it now. Silly me.
Many thanks. :)
View attachment 78203
Could you please explain that Fcos(theta) bit?
 
hjkchorong said:
Could you please explain that Fcos(theta) bit?
Nevermind, I've understood it now.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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