Calculating Applied Force in an Accelerating Elevator

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    Elevator Friction
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SUMMARY

The discussion focuses on calculating the applied force required to slide a box in an accelerating elevator. Given the upward acceleration of 1.84 m/s² and the total mass of the box at 28.9 kg, the normal force is calculated using the equation n = m(a + g), resulting in a normal force of 336.396 N. The force of kinetic friction is then determined using fk = μk * n, yielding a frictional force of 104.28 N. The final applied force must account for this friction to maintain constant speed.

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Homework Statement


You are riding in an elevator on the way to the eighteenth floor of your dormitory. The elevator's upward acceleration is a= 1.84 . Beside you is the box containing your new computer; box and contents have a total mass of 28.9 kg. While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door.

a)If the coefficient of kinetic friction between the box and elevator floor is uk= 0.31, what magnitude of force must you apply


Homework Equations


n=mgcosθ


The Attempt at a Solution


fy=n-mg=ma
n=m(a+g)=336.396
fk=un=0.31*336.396=104.28

I think this is the solution but I'm not entirely sure. As always any help is appreciated.
 
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Yes, looks good. But don't forget the proper units for the force.
 
thanks for te vote of confidence, I was a little worried I was wrong but I'm right thankfully :)
 

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