# Homework Help: Circular Motion of a car around a curve

1. Dec 29, 2013

### haha1234

1. The problem statement, all variables and given/known data

A car rounds a banked curve as discussed in Example 6.4 and shown in Figure 6.5.The radius of curvature of the road is R,the banking angle is θ,and the coefficient of static friction is μs.
Determine the range of speeds the car can have without slipping up or down the road.

2. Relevant equations

3. The attempt at a solution
I've just drawn two free-body diagram representing the situation that the centripetal acceleration is large that the car tends to sliding up and the opposite situation.
But I don't know how to find the range.

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Last edited: Dec 29, 2013
2. Dec 29, 2013

### Staff: Mentor

Good.

Apply Newton's 2nd law to each extreme. That will allow you to solve for the speeds.

3. Dec 29, 2013

### voko

What forces are there?

Observe that the net force on the car must be strictly horizontal.

4. Dec 29, 2013

### haha1234

Is the friction accounts for the centripetal acceleration?

5. Dec 29, 2013

### Staff: Mentor

What accounts for the centripetal acceleration is the net force on the car. That includes the friction.

6. Dec 29, 2013

### haha1234

There is my attempted solution:
tanθ=F/mg
F=mgtanθ
mgtanθ+fcosθ=mvmax2/R
gtanθ+μsg=vmax2/R
vmax=[Rg(tanθ+μs)]1/2
But it's not the correct answer.
Could you tell me the mistakes?

7. Dec 29, 2013

### Staff: Mentor

What is F?

Do this: List all forces acting on the car. There are three. Then write Newton's 2nd law for vertical and horizontal components.

8. Dec 30, 2013

### haha1234

Like this?

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9. Dec 30, 2013

### ehild

Where is the force of friction?

ehild

10. Dec 30, 2013

### Staff: Mentor

In your diagram, I'd like to see three vectors only--representing the three forces acting on the car.

Then I'd like you to list the horizontal and vertical components of each force. You'll use those components to write Newton's 2nd law.

11. Dec 30, 2013

### haha1234

How to find the horizontal component of the weight?
I think Wb in the picture is not parallel to the centripetal acceleration,so it may not be a component of the centripetal force.

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12. Dec 30, 2013

### ehild

W is vertical, it does not have horizontal component.

ehild

13. Dec 30, 2013

### haha1234

So is the weight giving the centripetal acceleration?

Last edited: Dec 30, 2013
14. Dec 30, 2013

### ehild

Why do you think so?

ehild

15. Dec 30, 2013

### Staff: Mentor

As ehild explained, since the weight acts vertically, the horizontal component will be zero.

For some reason, you found components of the weight parallel (Wb) and perpendiculur (Wa) to the incline surface. This was not necessary--stick to horizontal and vertical components.

(FYI: If you took the horizontal components of both Wa and Wb, you'll find that they cancel out.)

16. Dec 30, 2013

### haha1234

Finally I've found that the vmax=[Rg(tanθ+μs)]
But the right answer is [Rg(tanθ+μs)/(1-μstanθ)].
How (1-μstanθ) comes out?

17. Dec 31, 2013

### haruspex

You still have not clearly explained your working.
If N is the normal force and F the frictional force, what is the net force in the vertical direction, and what is the net force in the horizontal direction? (You can take F as being either up slope or down slope as you wish, so long as you are consistent.)

18. Dec 31, 2013

### haha1234

Is the net force in vertical direction 0N?
Is the net force in horizontal direction Nsinθ+fcosθ?

19. Dec 31, 2013

### Staff: Mentor

Again, don't just give answers, show your work step by step. Once again I request that you give the vertical and horizontal components of each force.

20. Dec 31, 2013

### Staff: Mentor

Yes, the net force in the vertical direction will equal 0, but express this as an equation: ∑Fy = 0. What is ∑Fy?

Yes, for one of the extremes.

21. Dec 31, 2013

### haha1234

Ny+W=0?

And I would like to know whether the sum of the the horizontal component of each force is the centripetal force.

22. Dec 31, 2013

### Staff: Mentor

There are three forces. Please list the horizontal and vertical components of each. Pay attention to sign.

That happens to be true, but the better way to understand it is to think of the acceleration and then apply Newton's 2nd law. In this case the acceleration is centripetal and it acts horizontally.

23. Dec 31, 2013

### haha1234

Vertical component:
Ny-W-fsin=0?

Horizontal component:
Nx+fcosθ=mv2/R?

24. Dec 31, 2013

### Staff: Mentor

Express Ny in terms of θ and N.

Same comment: Express Nx in terms of θ and N.

But you are getting closer! Also: Express the friction in terms of the normal force.

25. Dec 31, 2013

### haha1234

Nx+fcosθ=mvmax2/R
mgtanθ+μsNcosθ=mvmax2/R
gtanθ+μs(g/cosθ)cosθ=vmax2/R
vmax=[Rg(tanθ+μs)]1/2
But it's not the correct answer.