Question

1150 kg car is on a 8.70 degree hill. What’s the x component of the normal force

Answer 1

Answer:

The x - component of the normal force is equal to 1706.45 N.

Explanation:

To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:

$N

x

=W∗Sin(α)=mg∗Sin(α)$

$N

y

=W∗Cos(α)=mg∗Cos(α)$

Now, using the given information, we have:

$mass=m=1150Kg$

$a=8.70$

$g=9.81 \frac{m}{ {s}^{2} }$

Calculating, we have:

$N_{x}=mg*Sin(\alpha)$

$\: \: \: \: \: \: \: \: \: N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)$

$N

x =11281.5

s

2

Kg.m

∗Sin(8.70\°) = \\ \\ 1706.45

s

2

Kg.m

=1706.45.23N$

Hence, we have that the x-component of the normal force is equal to 1706.45 N.

Answer 2

Answer:

zero

Explanation:

fx=