Box sliding down wall at constant speed

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SUMMARY

The discussion centers on analyzing the forces acting on a 2.0 kg wood box sliding down a vertical wall at a constant speed while being pushed at a 45-degree angle. The primary equation used is ΣFY = FPY - mg, leading to the conclusion that the pushing force (FP) must counteract both gravity and friction. The correct calculation for FP includes the frictional force, represented as F(push)sin45 - μN - mg = 0, where μ is the coefficient of friction and N is the normal force (N = F(push)cos45). This adjustment is crucial for accurately determining the required pushing force.

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  • Knowledge of friction coefficients and their role in motion
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Homework Statement


The 2.0 kg wood box in the figure slides down a vertical wood wall while you push on it at a 45 ∘angle.
https://session.masteringphysics.com/problemAsset/1070374/4/06.P51.jpg

Homework Equations


ΣFY = FPY - mg

The Attempt at a Solution


I know that since there is a constant velocity, the acceleration will be zero. I then have
0 = FPsin45 - mg
FPsin45 = mg
FP = (mg)/sin45
FP = ((2kg)*(9.8m/s2))/sin45
FP = 27.719 N

Which is wrong. Am I correct in assuming that the only forces in the Y direction (taking up as positive Y) are the y-component of the pushing force and gravity? What am I missing here?
 
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Your approach just misses the friction component which acts against the relative motion of box/wall (i.e. downwards).. Therefore the equation will be

F(push)sin45 - mu*N - mg = 0 ; where mu is coefficient of friction and N= F(push) cos 45 ...
 

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