What is the horizontal component of the force pushing him forward?

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SUMMARY

The discussion centers on calculating the horizontal component of the force exerted by an 80-kilogram skier sliding on waxed skis at constant velocity. Key points include the understanding that since the skier is moving at constant velocity, the net force acting on him is zero, implying that the horizontal force exerted by the poles equals the frictional force, which is negligible in this scenario. The participants highlight the need for the coefficient of friction to solve the problem accurately, but conclude that for this simplified case, the horizontal force can be considered trivial or zero due to the absence of acceleration.

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  • Understanding of Newton's laws of motion
  • Knowledge of forces, including gravitational and normal forces
  • Familiarity with the concept of friction and coefficients of friction
  • Basic algebra for solving equations
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zelda1850
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Homework Statement



An 80 kilogram skier slides on waxed skis along a horizontal surface of snow at constant veloctiy while pushing with his poles. what is the horizontal component of the force pushing him forward?

2. Homework Equations [/]

is there an equation?

The Attempt at a Solution



i only know the mass how can i solve this problem?

this was mutiple choice

1) 0.05 N
2) 0.04 N
3) 40 N
4) 4N
 
Last edited:
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I think it's a trick question, seeing as he is skating (so friction is negligible) and moving at a constant velocity (no acceleration) he isn't exerting any more force, which seems to work conceptually, but the problem states he is actively moving the poles so he obviously is exerting a force so maybe it is a trick question...
 
oh but is there a way to solve this problem?
 
zelda1850 said:
oh but is there a way to solve this problem?
I suspect that you are given the coefficient of friction for waxed skis on snow. You'll need it.
 
Neglecting friction: F = m A. since acceleration is zero, the force is zero.

With friction this is a far more complex question than we are given data for. We would need to know not only the coefficient of friction for the skis, but also data on the poles such as the angle at which they push off the ground since this would lighten the force on the skis. Ideally this would be related as a periodic function of time.

I suspect the trivial answer is the one you're looking for.
 
DmplnJeff said:
With friction this is a far more complex question than we are given data for. We would need to know not only the coefficient of friction for the skis, but also data on the poles such as the angle at which they push off the ground since this would lighten the force on the skis. Ideally this would be related as a periodic function of time.
You are correct that a realistic answer would depend on the how the force of the poles is exerted and would be periodic. But this is meant as a simple problem, so just pretend that the poles only push backwards with no vertical component and that the force is continuously applied. Then the problem is easily solved (albeit naively).
 
im confused so i can use the equation F = M A to solve it? how can i calculate the answer?
 
What forces act on the skier? What's the net force on the skier?

Do you understand how to calculate the friction force? (You need the coefficient of friction.)
 
yeh u = ff/fn I am not sure what the net force on the skier is
 
  • #10
zelda1850 said:
im not sure what the net force on the skier is
What's the acceleration of the skier?
 
  • #11
the acceleration is at constant so its zero
 
  • #12
zelda1850 said:
the acceleration is at constant so its zero
The velocity is constant thus the acceleration is zero. So what does that tell you about the net force?
 
  • #13
the net force is zero?
 
  • #14
The forces on the skier include a gravitational force pushing down and a normal force opposing it. This normal force causes friction.
 
  • #15
The friction force and the force of the poles net to zero.
 
  • #16
zelda1850 said:
the net force is zero?
Good. Now answer my other question: What forces act on the skier?
 
  • #17
theres a normal force?
 
Last edited:
  • #18
The force of the snow pressing up on the skis. Otherwise the skier would fall to the center of the Earth.
 
  • #19
how can i find out the normal force of the skier?
 
  • #20
Unless something is being partially lifted (which doesn't apply to your problem) then normal force = force of gravity (mg)
 
  • #21
oh can someone just tell me how they solved it and explain it I am so confused
 

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