Circular motion problem (study guide help)

[PatrickL]

Homework Statement

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.

Homework Equations

Fnet=ma

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.

 

[Doc Al]

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?

 

[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.

 

[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.

 

[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.

 

[Doc Al]

I have a diagram and a freebody diagram

Good. Post it, if you can. Label all the angles that you need.

Heres what I know: I need to find the horizontal component of R

Before you find the horizontal component of R, first find R.

so do I use sine to do this or cosine?

You'll end up using both.

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?

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.

And is the 14 degree angle the angle I can use or do I need to find the other angle in the triangle?

You can do everything in terms of the 14 degree angle or the other angle. Realize that sinθ = cos(90 - θ).

I also know that the vertical component of the equation must equal 8500 Newtons since the car is not accelerating in the vertical direction.

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.

 

[swede5670]

Well I don't 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

 

[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

 

[PatrickL]

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

 

[Doc Al]

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

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?