Simple Physics Problem (pushing a crate up a slope)

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SUMMARY

The discussion focuses on calculating the minimum force required to prevent a 2.80 kg crate from sliding down a 35.0° incline, given a coefficient of static friction of 0.260. Participants analyzed the weight components, concluding that the normal force (N) is 15.75 N perpendicular to the slope. The correct calculation for the frictional force is 5.85 N, leading to a total force requirement of 38.1 N to maintain equilibrium. The initial miscalculation stemmed from incorrect component breakdown and misunderstanding of the forces involved.

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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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