Why Do We Use Sine for the X-Component in Physics Problems?

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Homework Help Overview

The discussion revolves around understanding the use of trigonometric functions, specifically sine and cosine, in resolving forces acting on a skier coasting down a slope. The original poster expresses confusion about why the sine function is applied to the x-component of the gravitational force in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the application of sine and cosine in the context of component forces, particularly questioning the use of sine for the x-component instead of cosine.
  • Some participants provide insights into the relationship between the forces and the geometry of the situation, suggesting that a free-body diagram is essential for clarity.
  • One participant mentions the nonconservative nature of friction and its implications for energy conservation in the problem.

Discussion Status

The discussion includes various interpretations of the problem, with some participants offering explanations that help clarify the original poster's confusion. However, there is no explicit consensus reached, as the original poster later indicates that their free-body diagram was incorrect, suggesting ongoing exploration of the topic.

Contextual Notes

The original poster expresses concern about their understanding of basic trigonometry in the context of physics, indicating a potential gap in foundational knowledge that may affect their ability to solve the problem. There is also mention of formatting issues with LaTeX, which may impact the clarity of shared equations.

Hemingway
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Homework Statement



I am worried that I don't understand a basic part of figuring out the component forces in the following problem. I have a full worked example but there is a few steps which I don't understand why we use sin for the x component and not cos (understand why I am really worried as it appears to be basic trig :/)

A 58-kg skier is coasting down a 25° slope, as Figure 6.7a shows. Near the top of the slope, her speed is 3.6 m/s. She accelerates down the slope because of the gravitational force, even though a kinetic frictional force of magnitude 71 N opposes her motion. Ignoring air resistance, determine the speed at a point that is displaced 57 m downhill.


Homework Equations


vf = √2(KEf) / m
= √(2(1/2 mv02 + Sigma F cos theta s)/m
= √(2(1/2 mv02 + mg sin 25 - fk s) /m
= √(2(1/2 mv02 + 170N cos 0 x 57) / 58
= 19m/s

The Attempt at a Solution



This was in my textbook:

a free-body diagram for the skier and shows the three external forces acting on her: the gravitational force , the kinetic frictional force , and the normal force . The net external force along the y-axis is zero, because there is no acceleration in that direction (the normal force balances the component mg cos 25° of the weight perpendicular to the slope). Using the data from the table of knowns and unknowns, we find that the net external force along the x-axis is:

SigmaF = mg sin 25 - fk
= (58)(9.8)(sin 25)(71)
= 170N


I look at sin and think we are looking at y component. I can't see why we would used sin for x component - can someone explain?

Many thanks

H
x

ps. please forgive formating - each time I use latex it just puts large gaps in the place of symbols :)
 
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AH! My free body diagram was incorrect - all good don't need a reply. thanks anyway those who read through my problem. Sorry for inconvenience.
 
Friction is a nonconservative force, so you can't use conservation of energy.

Sin is used for the x component because you have to draw a similar triangle for the normal force of the skiier. The angle between the force opposite the normal force and gravity is theta. Then trig will tell you that the x component is mgsin(theta).
 
Mindscrape said:
Friction is a nonconservative force, so you can't use conservation of energy.

Sin is used for the x component because you have to draw a similar triangle for the normal force of the skiier. The angle between the force opposite the normal force and gravity is theta. Then trig will tell you that the x component is mgsin(theta).

Thank you very much! This really helped me consolidate my understanding :)
 

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