What is the acceleration of a crate with two applied forces and friction?

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The discussion focuses on calculating the acceleration of a 25 kg crate subjected to two forces, F1 (88N at 55 degrees north of east) and F2 (54N east), while accounting for kinetic friction with a coefficient of 0.35. The initial calculations yielded an acceleration of 2.21 m/s² at 33 degrees north of east, but the book states the correct answer is 1.65 m/s² at 34 degrees north of east. The discrepancy arises from miscalculating the normal force and its relationship to the weight of the crate, which affects the frictional force applied.

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1. A 25 kg crate is initially at rest. Two forces, F1 and F2, are applied to the crate and it begins to move. The coefficient of kinetic friction between the crate and the floor is .35. Determine the magnitude and direction of (relative to the x-axis) the acceleration of the crate.

F1= 88N 55 degrees north of east.
F2= 54N east.

2. Equations:

Fk= (.35)Fn
a^2 + b^2= c^2

3. I worked out the problem and got an answer of 2.21 m/s^2 33 degrees north of east. However the answer in the book says 1.65 m/s^2 34 degrees north of east. I solved for the kinetic friction and got 85.75 N. I then broke the 88N and 85.75 friction down into horizontal and vertical components. I combined them, did the pythagorean thm, and found 55.75 for my force. I divided by 25kg to find the acceleration I got.

I did the problem again, but mistakingly switched the x and y components of the kinetic friction (49.18 and 70.242 respectively) and got the exact answer in the book? Could the book be wrong? Help please! Thanks.
 
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Hello and welcome to PF!

Your problem is that you treat the normal force as if it were equal to the weight of the box. Draw a free body diagram and you will see that if this were the case, the box would be accelerating upwards!
 
Sorry, I forgot to mention that the box is looked at from a "bird's-eye" perspective. Does that change anything?
 

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