# Circular motion problem (study guide help)

1. Dec 15, 2008

### PatrickL

1. The problem statement, all variables and given/known data
A car of weight 8500N is traveling at a constant speed along a road that is an arc of a circle. In order that the car may travel more easily round the arc, the road is banked at 14 degrees to the horizontal.
At one particular speed v of the car there is no frictional force at 90 degrees to the direction of travel of the car between the tires and the road surface. The reaction force of the road on the car is R (in the diagram R looks like normal force in a ramp problem).

a. Deduce that the horizontal component of the force R is approximately 2100N.
b. State the magnitude and direction of the resultant force acting on the car
c. Determine the speed v of the car at which it travels round the arc of radius 150 m without tending to slide.

2. Relevant equations
Fnet=ma
3. The attempt at a solution
I'm really confused about how to do this and I need it for my study guide so that I can start studying. I know that for A I have to use sine and cosine but I'm not sure how. I would really appreciate an explanation of how to do this, number A is the most important to understand (to quote my teacher) and I'm not even sure how to get it.

2. Dec 15, 2008

### Staff: Mentor

Start by drawing a free body diagram of the car on the incline. What forces act? Which way is the car accelerating? What must the sum of the vertical forces equal?

3. Dec 15, 2008

### rl.bhat

Draw a diagram,and identify the position of the car, inclination of the plane ,normal reaction and weight of the car. See whether you can get some idea.

4. Dec 15, 2008

### PatrickL

I have a diagram and a freebody diagram
Heres what I know: I need to find the horizontal component of R
so do I use sine to do this or cosine? Normally in these problems we've broken up the components of gravity and in that case the horizontal component has been found using sine(theta) x mg and the vertical component cos(theta) x mg

Can I do the same thing here? And is the 14 degree angle the angle I can use or do I need to find the other angle in the triangle?
I also know that the vertical component of the equation must equal 8500 Newtons since the car is not accelerating in the vertical direction.

5. Dec 15, 2008

### PatrickL

I'm going to need to end up substituting the two equations, I'm just not sure of the exact form that they're going to take. I appreciate that your trying to teach me how to do this, but right now I'd really like if you could at least tell me how to set up the equation. I'm spending a lot of time on this one problem and I have a lot of others in this packet that I have to get to. I don't mean to sound pushy, but I'm feeling really stressed out right now.

6. Dec 15, 2008

### Staff: Mentor

Good. Post it, if you can. Label all the angles that you need.
Before you find the horizontal component of R, first find R.
You'll end up using both.
In those cases gravity is a vertical force, which you're breaking into parallel and perpendicular components. So it's not quite the same thing, since R is perpendicular to the plane.
You can do everything in terms of the 14 degree angle or the other angle. Realize that sinθ = cos(90 - θ).
Good. That's the first step, which will allow you to calculate R.

To find the component of a vector X parallel to some direction, use Xcosθ where θ is the angle between X and the direction you want; for the perpendicular component, use Xsinθ. In this case, study the diagram to determine which is the angle R makes with the horizontal and vertical.

7. Dec 15, 2008

### swede5670

Well I dont know how to calculate the r as a whole I think I know how to get the component of r and that is the vertical component:
Cos(theta) * r = 8500. Is the angle 14 degrees? If it's not it must be 76 because all the angles must add up to 90
Hope that helps

8. Dec 15, 2008

### PatrickL

Ok. So I've made a triangle with r as the hypotenus and the adjacent (vertical component) is equal to 8500n. Since I have adjacent and an angle I can use cosine to find it right? Cos(14)=a/h and so cos (14) = 8500/h and then cos(14) * h = 8500 so 8500/cos(14)=h

9. Dec 15, 2008

### PatrickL

Can anyone help me understand if I'm heading in the right direction? If I am, does anyone know what B is asking

10. Dec 16, 2008

### Staff: Mentor

Good! So Rcos(14) = 8500, thus R = 8500/cos(14).

Now figure out the horizontal component of R.

Question b is asking about the net force on the car. The only forces acting on the car are the normal force (R, the reaction force) and gravity. What is their resultant?