Simple Physics Problem (pushing a crate up a slope)

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To prevent the 2.80 kg crate from sliding down a 35.0° incline with a static friction coefficient of 0.260, a minimum force must be applied perpendicular to the incline. The weight of the crate is broken down into components, with 15.75 N acting perpendicular and 22.5 N acting down the slope. The normal force is calculated to be 15.7 N, leading to a frictional force of 5.85 N. The total force needed to prevent sliding is miscalculated, as the correct answer is 38.1 N, indicating an error in component breakdown. A free-body diagram is suggested to clarify the forces at play.
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The coefficient of static friction between the m = 2.80 kg crate and the 35.0° incline of the figure below is 0.260. What minimum force must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

Attached is a picture.

I broke down the mass into 27.44 weight, which is not perpendicular to the slope so I broke down the 27.44 into components of 15.75 facing perpendicular towards the slope, and 22.5 going down the slope.

coefficient of static friction is .260(N) N= 15.7 since it is opposite of the weight.

so coefficient ends up being 4.1. so its Force should equal atleast 4.1+22.5. which is 26.6.

the answer is 38.1...what am I doing wrong here? Please help.
 
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it seems I broke down my components backwards, but I am still getting the wrong answer.

N=22.5, so friction is 5.85.

15.74+5.85= 21.59, not 38.1 :(
 
Did you draw a free-body diagram?
 
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