Question

A 1500 kg car drives at 30 m/s over a circular hill that has a radius of 370 m as shown in (Figure 1). At the point shown in figure 1, what is the normal force on the car?

Answer 1

9082 N

Step-by-step explanation:

Given a 1500 kg car traveling at 30 m/s over a circular hill with a radius of 370 m and positioned 30 degrees from the center of the circle, determine the normal force on the car.

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To determine the normal force on the car as it traverses a circular hill, we will compute the gravitational force acting downward and the "centripetal force" required for the car to maintain its circular motion. The difference between these two forces will give us the normal force.

Adding up the forces acting in the centripetal axis (refer to the attached image for a free-body diagram):

`\sum \vec F_c: \ \vec n \cos(\theta)+\vec w\cos(\theta)=m\vec a_c\\\\\\\\\Longrightarrow \vec n \cos(180 \textdegree)+\vec w\cos(30 \textdegree)=m\vec a_c\\\\\\\\\Longrightarrow\vec n \cos(180 \textdegree)+ mg\cos(30 \textdegree)=m\Big((v^2)/(r)\Big) \ \Big[\because \vec a_c=(v^2)/(r), \ \vec w =mg \Big]`

Solving the equation above for the normal force, 'n':

`\Longrightarrow\vec n \cos(180 \textdegree)+mg\cos(30 \textdegree)=m\Big((v^2)/(r)\Big)\\\\\\ \\\Longrightarrow -\vec n+mg\cos(30 \textdegree)=m\Big((v^2)/(r)\Big)\\\\\\\\\Longrightarrow mg\cos(30 \textdegree)=m\Big((v^2)/(r)\Big)+\vec n\\\\\\\\\therefore \vec n=m\Big(g\cos(30 \textdegree)-(v^2)/(r)\Big)`

Substitute in all the information we know:

`\Longrightarrow \vec n=(1500 \ kg)\Big((9.8 \ m/s^2)\cos(30 \textdegree)-((30 \ m/s)^2)/(370 \ m)\Big)\\\\\\\\\therefore \boxed{\vec n=9082 \ N}`

Thus, the normal force acting on the car as it goes over the circular hill is approximately 9082 N.