Kinetic friction of crate being pulled

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SUMMARY

The problem involves calculating the force P required to pull a 1.06×102 kg crate across a horizontal surface with a coefficient of kinetic friction of 0.193 at an angle of 28.9° above the horizontal. The initial calculation of the frictional force was correct at 200.5 N, but the error occurred in determining the force P. The correct approach requires dividing the frictional force by the cosine of the angle (28.9°), resulting in a required force P of approximately 220 N, not 229 N.

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jp04
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I've worked this problem an dget an answer, but when i enter it into our homework site, it says my answer of 229 N is wrong, do any of you see what I am doing wrong.
A 1.06×102 kg crate is being pulled across a horizontal floor by a force P that makes an angle of 28.9° above the horizontal. The coefficient of kinetic friction is 0.193. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

I found the force applied by friction to be 200.5 N (F= (0.193)(106kg)(9.8m/s2)

so to find the force of P i said the force in the opposite x direction needs to be equal to the frictional force

cos28.9= 200.49/P

=229 N


If you see anything I'm doing wrong or another way to approach it tell me.

I don't think this is suppose to be a difficult problem.

Thanks in advance!
 
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jp04 said:
I've worked this problem an dget an answer, but when i enter it into our homework site, it says my answer of 229 N is wrong, do any of you see what I am doing wrong.
A 1.06×102 kg crate is being pulled across a horizontal floor by a force P that makes an angle of 28.9° above the horizontal. The coefficient of kinetic friction is 0.193. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

I found the force applied by friction to be 200.5 N (F= (0.193)(106kg)(9.8m/s2)

so to find the force of P i said the force in the opposite x direction needs to be equal to the frictional force

cos28.9= 200.49/P

=229 N


If you see anything I'm doing wrong or another way to approach it tell me.

I don't think this is suppose to be a difficult problem.

Thanks in advance!
In your calculation of the frictional force you have neglected the vertical component of P. The normal force is the weight of the crate, less the vertical component of P.

Draw a free body diagram with ALL of the forces acting on the crate. Then add separately all the vertical components and all the horizontal components.
 


Your approach to finding the force of P is correct. However, the answer of 229 N is incorrect. To find the magnitude of P, you need to divide the frictional force by the cosine of the angle between P and the horizontal direction. In this case, the angle is 28.9°, so you should divide by cos(28.9°) instead of cos(28.9). This will give you a force of approximately 220 N. Make sure to double check your calculations and units to ensure accuracy.
 

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