Normal Force and Static Frictional Force

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SUMMARY

The discussion focuses on calculating the normal force and static frictional force for a car parked on a 14.7° incline. The car's mass is 1710 kg, leading to a gravitational force of 16758 N. The normal force is determined to be 4252.48 N using the equation F_N = mg cos(θ), and the static frictional force, calculated using the coefficient of static friction (1.0), is also 4252.48 N. The user seeks clarification on entering these values with three significant digits for an online submission.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of trigonometric functions (sine and cosine)
  • Familiarity with the concept of static friction and its coefficient
  • Ability to perform calculations involving forces and angles
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  • Study the derivation of normal force on inclined planes
  • Learn about the application of static friction in real-world scenarios
  • Explore advanced problems involving multiple forces on inclined surfaces
  • Review the significance of significant figures in scientific calculations
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Students studying physics, particularly those focusing on mechanics and forces, as well as educators looking for examples of static friction and normal force calculations.

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Homework Statement



A car (m = 1710 kg) is parked on a road that rises 14.7 ° above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?

Coefficient of static friction: rubber on dry concrete: 1.0


Homework Equations



F=ma
Fs max = coefficient of friction x normal force

The Attempt at a Solution



I drew some pictures, but I'm slightly computer illiterate so..sorry. Anyway, not really sure, but here's what I got:

F= (1710 kg)(9.8 m/s^2)
F= 16758 N

Attempt at solution of the normal force:

16758 N x sin (14.7) = 4252.48 N

Static Frictional Force:

(I'm really not sure of this, but I'll go ahead)

(4252.48 N)(1.0)= 4252.48


I'm just at my wit's end. I've been trying to figure out this problem for two hours straight. I have to enter my answers in online, and I can only have three significant digits, yet I can't enter in scientific notation, so...I just don't know! Help please!
 
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You're thinking too hard about it. If you figure out the sum of forces you have

<br /> \begin{align*}<br /> \sum F_x &amp;= f_s - mg\sin\theta = 0\\<br /> \sum F_y &amp;= F_N - mg\cos\theta = 0<br /> \end{align*}<br />​
 

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