Basic force with static friction problem

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SUMMARY

The discussion centers on calculating the resultant downhill force acting on a 70-kg skier on a 20-degree slope, considering a 15 N air resistance force and a friction coefficient of 0.09. The skier's weight is approximately 700 N, derived from the mass and gravitational acceleration (70 kg x 10 m/s²). To solve the problem, one must break down the forces into their x and y components, using the equation for force, which is the product of the friction coefficient and the normal force.

PREREQUISITES
  • Understanding of basic physics concepts such as force, weight, and friction.
  • Knowledge of vector decomposition to analyze forces in two dimensions.
  • Familiarity with trigonometric functions, particularly in right triangles.
  • Ability to apply Newton's laws of motion in practical scenarios.
NEXT STEPS
  • Study the principles of vector decomposition in physics.
  • Learn how to calculate normal force in inclined planes.
  • Investigate the effects of friction on motion, specifically static and kinetic friction.
  • Explore the application of Newton's second law in real-world problems.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of motion on inclined surfaces, particularly in the context of sports like skiing.

Ering
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Homework Statement



A 70-kg skier is in a tuck position and moving down a 20-degree hill. Air resistance applies a resultant force of 15 N against the movement of the skier. The coefficient of friction between the skis and the snow is 0.09. What is the resultant downhill force acting on the skier?

Homework Equations



Force = friction coefficient x normal force

The Attempt at a Solution



I know you have to break down the force into its x and y components. So making a right triangle with the weight (~700 N: 70kg x 10 m/s2) as the hypotenuse and 70 degrees (90 - the given 20 degrees) as the angle. But after doing that, I'm not sure where to go from there. Is the resultant force also your x-component in the right triangle?
 
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Ering said:
Is the resultant force also your x-component in the right triangle?
The resultant force in a given direction is the sum of all the components that are in that direction of the applied forces. What applied forces have components in the direction of interest?
 

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