Kinetic friction of crate being pulled

AI Thread Summary
The problem involves calculating the force P required to pull a crate while considering kinetic friction. The user initially calculated the frictional force as 200.5 N but failed to account for the vertical component of force P, which affects the normal force. To find the correct magnitude of P, it is necessary to divide the frictional force by the cosine of the angle (28.9°), leading to an approximate value of 220 N instead of 229 N. A free body diagram is recommended to visualize all forces acting on the crate. Accurate calculations and unit checks are essential for obtaining the correct answer.
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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