When can you assume that gravity is the only applied force opposing friction?

Fnet(x) = n - W(x) = ma --> n - Wcos(theta) = maFnet(y) = f(k) - W(y) = 0 --> f(k) - Wsin(theta) = 0The normal force in that direction is equal to the weight of the skier.f
  • #1
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When can you assume that gravity is the only "applied force" opposing friction?

Homework Statement


"Determine the stopping distance for a skier moving down a slope with friction with an initial speed of 24.7 m/s. The slope makes an angle of 5.02deg above the horizontal, and assume that μk=0.172."

For this question, I have three variables that I don't have values for -- normal, mass, and acceleration. Since I only have two equations (sum of the forces for F(x) and F(y) directions), I can only have two unknowns for me to solve the problem. Am I supposed to assume that the acceleration is 9.8m/s/s (acceleration due to gravity) in this case?


Homework Equations


Fnet = ma


The Attempt at a Solution


Fnet(x) = n - W(x) = ma --> n - Wcos(theta) = ma
Fnet(y) = f(k) - W(y) = 0 --> f(k) - Wsin(theta) = 0

With my method of assuming that a=9.8m/s/s, I then isolated for "m" in each equation.
(1) m = n / [a +gcos(theta)]
(2) m = μ*n / [gsin(theta)]

I then equated both formulas to each other, to solve for the normal, "n".
 
  • #2


For this question, I have three variables that I don't have values for -- normal, mass, and acceleration.
You might not need the mass--just call it "m" and continue. The normal force can be expressed in terms of the mass.
Since I only have two equations (sum of the forces for F(x) and F(y) directions), I can only have two unknowns for me to solve the problem. Am I supposed to assume that the acceleration is 9.8m/s/s (acceleration due to gravity) in this case?
Definitely not! The acceleration is what you need to find in order to solve for the stopping distance.

The Attempt at a Solution


Fnet(x) = n - W(x) = ma --> n - Wcos(theta) = ma
Looks like you chose the x-coordinate to be perpendicular to the surface. What's the acceleration in that direction?
Fnet(y) = f(k) - W(y) = 0 --> f(k) - Wsin(theta) = 0
Parallel to the surface the acceleration is not zero. That's what you're trying to find.

Usually folks take the x-axis as parallel to the surface, but it doesn't matter. The resulting equations are what matters.

Express the weight, normal force, and friction force in terms of the mass. (And other needed values, of course.)
 
  • #3


You might not need the mass--just call it "m" and continue. The normal force can be expressed in terms of the mass.

From the previous post, I just need some clarification on what it means to express the normal force in terms of the mass.

I know that on a non-inclined surface, normal = weight = mass * gravity... but this is an inclined plane. I thought we couldn't assume that normal = m*g?
 
  • #4


I know that on a non-inclined surface, normal = weight = mass * gravity... but this is an inclined plane. I thought we couldn't assume that normal = m*g?
Right, you can't assume that. Use your equation for Fnet for forces perpendicular to the surface. You'll have to correct it. What must Fnet equal in that direction?
 

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