Kinetic friction of crate being pulled

In summary, the conversation discusses a problem involving a crate being pulled across a horizontal floor. The person has calculated the force of friction to be 200.5 N and is trying to find the magnitude of the pulling force P. However, when they enter their answer of 229 N into the homework site, it is marked incorrect. They ask for guidance and someone suggests taking into account the vertical component of P. The conversation ends with the person expressing their belief that the problem should not be difficult.
  • #1
jp04
2
0
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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  • #2
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.
 
  • #3


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.
 

FAQ: Kinetic friction of crate being pulled

What is kinetic friction?

Kinetic friction is the force that opposes the motion of an object when it is in motion.

What factors affect the kinetic friction of a crate being pulled?

The factors that affect the kinetic friction of a crate being pulled include the weight of the crate, the type of surface the crate is being pulled on, and the force applied to the crate.

How is the coefficient of kinetic friction calculated?

The coefficient of kinetic friction is calculated by dividing the force of kinetic friction by the normal force acting on the object.

Is the kinetic friction of a crate being pulled constant?

No, the kinetic friction of a crate being pulled is not constant. It can change depending on the factors mentioned above.

How can the kinetic friction of a crate being pulled be reduced?

The kinetic friction of a crate being pulled can be reduced by using a lubricant, reducing the weight of the crate, or using a smoother surface to pull the crate on.

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