Skier traveling uphill with and without friction

In summary: Correct. The component perpendicular to the slope is mg*sin(30).Correct. The component perpendicular to the slope is mg*sin(30).In summary, the problem asks for the distance up the hill a skier travels given a coefficient of kinetic friction of 0.11 and a slope of 30 degrees. The solution involves considering the forces of friction and gravity and their effects on the skier's motion. Using trigonometry, the gravitational force can be broken
  • #1
m_eghan
19
0

Homework Statement


If the coefficient of kinetic friction in the previous problem was actually 0.11 and the slope 30 degrees, to the nearest meter how far up the hill does he go?

**the problem prior was: A skier traveling at 31.9 m/s encounters a 12 degree slope. If you could ignore friction, to the nearest meter, how far up the hill does he go? (my answer was 250m up the hill)**

Homework Equations


Sum of the forces in the y-axis = Fn - Fgy = 0
Sum of the forces in the x-axis = Ff + Fgx = ma

The Attempt at a Solution


I figured acceleration = Fnet / mtotal
therefore, acceleration = ((.14*mg*cos30)+(mgsin30)) / m
then the m's would cancel and it would just be..
a = (.14*g*cos30)+(gsin20)
 
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  • #2
... and, your question is what?
 
  • #3
Bystander said:
... and, your question is what?
well what I thought the solution was, was incorrect. so I was looking for some guidance.
 
  • #4
What forces contribute to deceleration of the skier?
 
  • #5
Bystander said:
What forces contribute to deceleration of the skier?
Just friction
 
  • #6
m_eghan said:
coefficient of kinetic friction in the previous problem was actually 0.11

m_eghan said:
and the slope 30 degrees
... and, what else?
 
  • #7
Bystander said:
... and, what else?
gravity?
 
  • #8
Friction and gravity. Correct. In what directions are these forces acting?
 
  • #9
Bystander said:
Friction and gravity. Correct. In what directions are these forces acting?
friction is opposite of the skiers motion, parallel to the surface & gravity is straight down
 
  • #10
The surface is inclined 30° to the horizontal; how does this affect the gravitational force?
 
  • #11
Bystander said:
The surface is inclined 30° to the horizontal; how does this affect the gravitational force?
It increases it...?
 
  • #12
I'll give you a conditional "yes:" if the inclination to horizontal is 0, what is the gravitational force? and, if the inclination is 90°, what is the gravitational force?
 
  • #13
Wat is the correct answer?
I get it as 87.13m
m_eghan said:

Homework Statement


If the coefficient of kinetic friction in the previous problem was actually 0.11 and the slope 30 degrees, to the nearest meter how far up the hill does he go?

**the problem prior was: A skier traveling at 31.9 m/s encounters a 12 degree slope. If you could ignore friction, to the nearest meter, how far up the hill does he go? (my answer was 250m up the hill)**

Homework Equations


Sum of the forces in the y-axis = Fn - Fgy = 0
Sum of the forces in the x-axis = Ff + Fgx = ma

The Attempt at a Solution


I figured acceleration = Fnet / mtotal
therefore, acceleration = ((.14*mg*cos30)+(mgsin30)) / m
then the m's would cancel and it would just be..
a = (.14*g*cos30)+(gsin20)
 
  • #14
Bystander said:
I'll give you a conditional "yes:" if the inclination to horizontal is 0, what is the gravitational force? and, if the inclination is 90°, what is the gravitational force?
If i's 0, the Gravitational force is equal to mass*9.8. If it's 90, I'm not exactly sure
 
  • #15
Think again, if the skier hits a horizontal stretch, how much effect does gravity have on his velocity?
 
  • #16
Bystander said:
Think again, if the skier hits a horizontal stretch, how much effect does gravity have on his velocity?
Oh, none because Gravity is in the y-axis and the skiers motion is in the x... So, does that mean at 90° gravity would cause him to decelerate?
 
  • #17
Yes. Now, by how much does gravity decelerate as a function of the surface inclination?
 
  • #18
Yes . the gravity ,actusly it's sin component will cause him to decelerate. Besides he is not moving in x axis.
m_eghan said:
Oh, none because Gravity is in the y-axis and the skiers motion is in the x... So, does that mean at 90° gravity would cause him to decelerate?
 
  • #19
Bystander said:
Yes. Now, by how much does gravity decelerate as a function of the surface inclination?
Would it be -9.8?
 
  • #20
Tip the surface 1 degree at a time from horizontal to vertical; for each increment of increase in the slope, how much does the effect of gravity increase?
 
  • #21
Bystander said:
Tip the surface 1 degree at a time from horizontal to vertical; for each increment of increase in the slope, how much does the effect of gravity increase?
for 1 degree, it's effected by cos(1), then for 2 degrees it's cos(2) etc.
 
  • #22
If it's zero for horizontal, which trig function do you want to use?
 
  • #23
Bystander said:
If it's zero for horizontal, which trig function do you want to use?
I'm thinking it's probably tangent, but I have no idea why
 
  • #24
Cosine, tangent, four more guesses. Diagram it: we'll move the skier from left to right up the 30 degree slope; gravitational force is mg straight up and down; the two components you're interested in are 1) parallel to the slope and 2) perpendicular (normal) to the slope. mg is the hypotenuse of a right triangle, the component parallel to the slope can be called the base, and the component normal to the slope something else. Where is the 30 degree interior angle of the right triangle we just sketched?
 
  • #25
Bystander said:
Cosine, tangent, four more guesses. Diagram it: we'll move the skier from left to right up the 30 degree slope; gravitational force is mg straight up and down; the two components you're interested in are 1) parallel to the slope and 2) perpendicular (normal) to the slope. mg is the hypotenuse of a right triangle, the component parallel to the slope can be called the base, and the component normal to the slope something else. Where is the 30 degree interior angle of the right triangle we just sketched?
https://mail.google.com/mail/u/0/?ui=2&ik=af651a9087&view=fimg&th=14a443a0b04882eb&attid=0.1&disp=inline&safe=1&attbid=ANGjdJ8jLgbV6GQoMneNUtdYs18obcX3DiSHF3TW-aeKwBQ1IJjf6oW3QpAXkq4Ja1j2DaU87wb2S-Ve7MCeBT7i1xWYToQoJo3g8Bm1RUo6pw1c9UISprC_Xf8j2RU&ats=1418483866321&rm=14a443a0b04882eb&zw&sz=w1342-h547
 
  • #26
m_eghan said:
E]
[PLAIN]https://gm1.ggpht.com/yJ098TyA21RmFpkD2D0lDipFpTHoFUxXntJ7lJ6A0o6R-HgUEPcZLUB8uQQFZF07NjgRHSG0u6hlIWWThZ8T-TcfEhTUtce_VPVe3awBfQ4GBqKvFW9B2o_iSaEtYlt9FQZ8hjAen3FlnhrtwdGcwU3gbvfr5epvZiOqbtjOlpQU3ErQAE-xtOhmqewjAKVVFuzAp9kdmShr9ysUvrbafQfBJd-QR76wh763VKwIePBoNHrywFCOfGYBgEVGPe31sDX1x841USziaaoYYlrxi4XMwGUSezv8AuV0YYz8XYtkg1II9tAbqX87oTtUmeaKf_KBJL-2E5nREtd19FLSm9GnA32diY2r1wv1PAGSW4h3ppwFcnssbeD1Wc6aFwhEBiSeBbMWZyxVWD77B02tLN5OQdgiMjKE1lvznEnUUEKcWrItMuBXfCsTkWeT08xU1hPOJRZqhN1ZUG_uTL9cohmhPD02jGzr0VKeO9PXGFj-lXgzc-Y3O0_q1LPfw1EN_quOk5ddgLL_Vxjeg9Gr_TTBi6rrz0_RbDOFGSDQn9tOxrM_6GtiX_IRITytpgvux1UyL50=w1342-h547-l75-ft[/QUOTE]
 
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  • #27
You sketched it to match your thinking rather than mine, which is fine. Now, you want the component parallel to your x-axis, and that is mg times "what?"
 
  • #28
Bystander said:
You sketched it to match your thinking rather than mine, which is fine. Now, you want the component parallel to your x-axis, and that is mg times "what?"
mg * sin30
 
  • #29
Bingo. And, the normal force? That's the component parallel to the y-axis.
 
  • #30
Bystander said:
Bingo. And, the normal force? That's the component parallel to the y-axis.
mg * cos30
 
  • #31
Then the total force decelerating the skier is ... ?
 
  • #32
Bystander said:
Then the total force decelerating the skier is ... ?
mg*cos30 + mg*sin30 ??!??
 
  • #33
You forgot the coefficient of friction.
 
  • #34
Bystander said:
You forgot the coefficient of friction.
OHHH!
μ*mg*sin30 + mg*cos30
 
  • #35
Close. Frictional force is the product of coefficient of friction and the normal force. One more time.
 

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