smashbrohamme
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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.
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.