Find the frictional force on the skis

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Homework Help Overview

The discussion revolves around a physics problem involving a skier descending a 10° slope at constant velocity. The skier's mass is 57 kg, and the force of air resistance is given as 74 N. Participants are tasked with determining the frictional force acting on the skis.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the normal force and its calculation, with one noting a value of 558 N. There are attempts to analyze the forces acting on the skier, including air resistance and gravitational components. Questions arise regarding the direction of these forces and how they relate to the slope.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces involved. Some guidance has been offered regarding the components of forces acting parallel and perpendicular to the slope, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants are grappling with the definitions of forces and their components, particularly in relation to the slope. There is mention of drawing a free body diagram (FBD) to clarify the forces, but some participants express confusion about the setup and calculations.

ramenluver50
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Homework Statement


A skier of mass 57 kg skis down a 10° slope @ constant velocity.
The force of air resist. has a magnitude of 74 N. Find the frictional force on the skis.


Homework Equations



F=ma
F(fr) = (mu)(Fn)


The Attempt at a Solution



I found Fn= ma
where is 558N
so far i tried, 74sin10- 558
and i tried drawing a body diagram, where 558 is directly upwards, 9.8 down for gravity, the downward slope of 74 N at 10 deg
 
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ramenluver50 said:

Homework Statement


A skier of mass 57 kg skis down a 10° slope @ constant velocity.
The force of air resist. has a magnitude of 74 N. Find the frictional force on the skis.


Homework Equations



F=ma
F(fr) = (mu)(Fn)


The Attempt at a Solution



I found Fn= ma
where is 558N
so far i tried, 74sin10- 558
and i tried drawing a body diagram, where 558 is directly upwards, 9.8 down for gravity, the downward slope of 74 N at 10 deg

Hint -- sum the forces in the direction of the motion of the skier (along the direction of the slope). What do those forces need to sum to?
 
it would be the sum of normal force, and air resistance?
so it should be 74 +558 ?
 
ramenluver50 said:
it would be the sum of normal force, and air resistance?
so it should be 74 +558 ?

The normal force is orthogonal to the slope, not parallel to it. Which way does the air resistance vector point? The friction force vector? What force vector opposes those two vectors?
 
air resistance is opposite to the velocity, and friction force should be the same,

gravitainal force opposes it but i don't think directly bc its is not parallel to the slope.
 
ramenluver50 said:
air resistance is opposite to the velocity, and friction force should be the same,

gravitainal force opposes it but i don't think directly bc its is not parallel to the slope.

But one component of the gravitational force is parallel to the slope... Is it the sin or cos component? Remember that if the slope is zero degrees, the gravitational component of force paralle to the slope is zero...
 
it is sin, (lookin for the Y component of the slope)

which i had (74 sin 10),
 
ramenluver50 said:
it is sin, (lookin for the Y component of the slope)

which i had (74 sin 10),

74 N is the force from air resistance, not gravity.

Write the sum of the forces in the direction parallel to the slope. What do you want to set that sum to equal, and why?
 
umm I am lost now... maybe i drew it wrong...

the air resistance is the magnitude isn't it?

iono.. start over?
 
  • #10
ramenluver50 said:
umm I am lost now... maybe i drew it wrong...

the air resistance is the magnitude isn't it?

iono.. start over?

Describe (or post) your FBD. What force vectors act on the skier?
 
  • #11
what is FBD? , the forces affecting the skier is air resistance, *frictional force, gravity, velocity
 
  • #12
FBD = free body diagram. You alluded to it in your first post.

And velocity is not a force...
 
  • #13
down is 9.8 (gravity,) down slope is 74 in mag. the slope from the X axis is 10 deg from it.the Y of the triangle i believe is (74 sin 10) and perpendicular to the slope i have upward is 558 N
 

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