Frictional force/coeff of friction

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In summary, the skier has a mass of 64kg, starts at the top of a 18 degree slope and let's gravity take her downhill. The slope is 65 meters long and her speed is 15 meters per second when she get to the bottom.To find the x component of weight, the skier needs to find the coefficient of friction and the frictional force. The coefficient of friction is .14 and the frictional force is 83N.
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pizat98
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I have been working at this question for a while. I have to find the frictional force and the coefficient of friction. skier has a mass of 64kg, starts at top of a 18 degree slope and let's gravity take her downhill. slope is 65 m long and her speed is 15 m/s when she get to the bottom. how do I find the x component of weight as well? The answer for coefficient is 0.14N according to the book and frictional force is 83N.



Fnet=mg
fk=ukFn




I came up with Fn=605N in relation to the y axis. a=3.03 m/s^2 (alteast what I got)
I have tried a thousand different ways to do this and I keep getting myself more confused. The number I keep coming up with for coeff is .32 and I don't have a clue about frictional force. Any help would be great.
 
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  • #2
My first suggestion is to draw a free-body diagram if you have not done so yet. Next, write out the sum of the forces in the x and y directions.
 
  • #3
I have written out the free body diagram. the sum of the forces for x I have is 197N and for y it is 605. I am not sure if these are exactly correct. for y I used mg cos theta, and for x I used mg sin theta
 
  • #4
Assuming you've got your coordinate axes oriented properly, you should find that the net force in the y direction is 0. As a suggestion, don't plug numbers in until you arrive at your final expression.

Some help since I will be signing off for the night:

Fy = F_N - mgcos(theta) = 0

Fx = mgsin(theta) - F_f = ma

Therefore, F_f = mgsin(theta) - ma

Since F_f is of constant magnitude, the acceleration will be uniform, hence, V_f^2 = 2ax.

Hopefully you can do the rest from here on out, good luck.

P.S. I think the answer for the friction force should be 87 N, I'm not positive though, my math could be wrong.
 
  • #5
Thanks very much. I got 85 for my frictional force and .14 for coeff. The book says 84N for frictional and .14 for coeff. That should be close enough. I am off one for frictional force, but that could be rounding errors. Thanks again
 

What is frictional force?

Frictional force is the force that resists the relative motion between two surfaces in contact. It is caused by the microscopic roughness of the surfaces and the interlocking of the irregularities when they come into contact.

What is the coefficient of friction?

The coefficient of friction is a dimensionless quantity that represents the ratio of the frictional force between two surfaces to the normal force pressing them together. It is a measure of how easily two surfaces slide against each other.

What factors affect the coefficient of friction?

The coefficient of friction is affected by several factors, including the nature of the surfaces in contact, the roughness of the surfaces, the applied force, and the presence of any lubricants or contaminants. It also depends on the temperature and humidity of the environment.

How is frictional force calculated?

Frictional force can be calculated by multiplying the coefficient of friction by the normal force between two surfaces. The normal force is the force perpendicular to the surfaces in contact. This calculation assumes that the surfaces are not moving, and the coefficient of friction remains constant.

How can frictional force be reduced?

Frictional force can be reduced by using lubricants between the surfaces, making the surfaces smoother, or using materials with lower coefficients of friction. Additionally, reducing the force pressing the surfaces together can also decrease the frictional force. For example, in car engines, oil is used as a lubricant to reduce friction between moving parts and increase efficiency.

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