What is the relationship between force and surface area when pushing a crate?

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When pushing a crate across the floor, the force required to maintain constant speed is influenced by the friction force, which is calculated using the equation F_friction = μ_k F_normal. Reducing the surface area in contact with the floor does not affect the friction force, as it is proportional to the weight of the object rather than the contact area. Therefore, even with half the surface area, the force needed to push the crate remains the same as before. This highlights that friction is dependent on the normal force and the coefficient of friction, not the area of contact. Understanding this relationship is crucial for accurately predicting the force needed to move objects.
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You are pushing a wooden crate across the floor at constant speed.You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be what compared to the original force?
 
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This can't be that hard
 
Depending on your level, force of friction is using a simple model calculated by the following:
<br /> F_\textrm{friction} = \mu_k F_\textrm{normal}<br />
Keep in mind this isn't a fundamental law, it's an experimental relation between the magnitude of the friction force and the magnitude of the normal force.
 
Remember that the friction force is proportional to the weight of the object, not the area of contact!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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