# Free-body diagram to identify the forces acting on the car

1. Oct 21, 2004

### kimikims

Where do I start?

A curve of radius 56.5 m is banked so that
a car traveling with uniform speed 59 km/hr
can round the curve without relying on friction
to keep it from slipping to its left or right.
The acceleration of gravity is 9.8 m/s^2

What is Q? Answer in units of degrees.

2. Oct 21, 2004

### Sirus

Again, use a free-body diagram to identify the forces acting on the car. Remember, the car is travelling in circular motion; what does that tell you?

3. Oct 23, 2004

### kimikims

Is this the right equation?

Mg = 1.3
g = 9.8 m/s^2
Speed = 59km/hr
m = 0

Mg Tan = mV^2/R

1.3 Mg Tan = 0 x (59)^2 / 56.5

But wouldn't that just make it zero?? I'm lost.

4. Oct 23, 2004

### arildno

What is M and what is m??
and why is m=0???????????????????

5. Oct 23, 2004

### kimikims

Wait...

M x g tan = m (V^2/g)

g tan = V^2 / g

9.8 tan = [(16.2)^2 / 56.5]

9.8 tan = 4.644955752 (Do I just type it in as that? or do I need to divide by 9.8)

I am trying to find the degree of the angle.

6. Oct 23, 2004

### thermodynamicaldude

k...in this case....the force of gravity = centripital force....which I think you have figured out already....

You are correct in stating that Centripital force= mv^2 / R

Now draw a free body diagram to find the force of gravity.....its not mg tan (angle)....its mg * something......

...work from there.

7. Oct 23, 2004

### kimikims

tan = V^2 / Rg ??

tan = (59)^2 / (56.5) (9.8)

tan = .62867905

= .0109729727 ?

that doesnt make sense

8. Oct 23, 2004

### Pyrrhus

Kimikins,

$$\tan \theta =\frac{v^2}{rg}$$

Only works without friction!, it's the formula to find the angle of a frictionless banked curve. To solve this problem do as the other says, do a freebody diagram and identify all the forces (components) pointing toward the center.