Question

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

Answer 1

zero

Step-by-step explanation:

fx=

Answer 2

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

Step-by-step 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 </p><p>x</p><p> </p><p> =W∗Sin(α)=mg∗Sin(α)`

`N </p><p>y</p><p> </p><p> =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(m)/(s^(2))*Sin(8.70\°)`

`N </p><p>x =11281.5 </p><p>s </p><p>2</p><p> </p><p>Kg.m</p><p> </p><p> ∗Sin(8.70\°) = \\ \\ 1706.45 </p><p>s </p><p>2</p><p> </p><p>Kg.m</p><p> </p><p> =1706.45.23N`

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