Theoretical situation - surface area vs. friction

[LadyMario]

You are pushing a wooden crate across the floor at a constant speed. You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be about: A) four times as great B) twice as great C) equally great D) half as great E) one fourth as great

I'm pretty sure surface area does not relate to the force needed; as friction equations don't account for surface area just mass. But otherwise, I'm really not sure, and there must be more reason to it than that the simple Newton's laws equations not accounting for surface area...

 

[Dick]

If you think in terms of area you could describe the frictional force as the coefficient of friction times the normal force per unit area (pressure) times the area. But when you halve the area, you also double the normal force per unit area. So the frictional force is unchanged. That's why you don't need to consider the area. Does that help??

 

[Chestermiller]

You are pushing a wooden crate across the floor at a constant speed. You decide to turn the crate on end, reducing by half the surface area in contact with the floor. In new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be about: A) four times as great B) twice as great C) equally great D) half as great E) one fourth as great

I'm pretty sure surface area does not relate to the force needed; as friction equations don't account for surface area just mass. But otherwise, I'm really not sure, and there must be more reason to it than that the simple Newton's laws equations not accounting for surface area...

On a more fundamental level, in terms of stress, the shear stress at the interface is equal to normal stress times the coefficient of friction. For the case in which the crate is turned on end, what happens to the contact area? What happens to the normal stress? What happens to the shear stress? What happens to the shear force?

 

[LadyMario]

If you think in terms of area you could describe the frictional force as the coefficient of friction times the normal force per unit area (pressure) times the area. But when you halve the area, you also double the normal force per unit area. So the frictional force is unchanged. That's why you don't need to consider the area. Does that help??

Very much so! Thank you. I knew the answer; just couldn't explain why

 

[Nickie]

I would say that the force required to push the crate across the floor at a constant speed would not change significantly with the reduction in surface area. This is because friction, which is the force that opposes motion, is not directly affected by surface area. It is determined by the coefficient of friction, which is a property of the two surfaces in contact.

In this theoretical situation, the force required to push the crate would likely be slightly less due to the reduction in surface area, but it would not be a significant difference. It is possible that the force required could be slightly more due to the change in the distribution of weight on the crate, but this would also be a small effect.

Therefore, the correct answer to the question would be C) equally great. The force required to push the crate would be approximately the same regardless of the orientation of the crate.