Normal Force and Static Frictional Force

AI Thread Summary
The discussion revolves around calculating the normal force and static frictional force for a car parked on a 14.7° incline. The weight of the car is calculated to be 16,758 N, and the normal force is derived using the sine of the angle, resulting in approximately 4,252.48 N. The static frictional force is then calculated using the coefficient of static friction, yielding the same value of 4,252.48 N. The user expresses frustration with the problem-solving process and the requirement for significant digits without scientific notation. Clarification on the sum of forces in both x and y directions is suggested to aid in understanding the problem.
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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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