Circular Motion of a car around a curve

[haha1234]

Homework Statement

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 μ_s.
Determine the range of speeds the car can have without slipping up or down the road.

Homework Equations

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.

 

[Doc Al]

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.

Good.

But I don't know how to find the range.

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

 

[voko]

What forces are there?

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

 

[haha1234]

Good.


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

Is the friction accounts for the centripetal acceleration?

 

[Doc Al]

Is the friction accounts for the centripetal acceleration?

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

 

[haha1234]

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

There is my attempted solution:
tanθ=F/mg
F=mgtanθ
mgtanθ+fcosθ=mv_max²/R
gtanθ+μ_sg=v_max²/R
v_max=[Rg(tanθ+μ_s)]^1/2
But it's not the correct answer.
Could you tell me the mistakes?

 

[Doc Al]

There is my attempted solution:
tanθ=F/mg
F=mgtanθ

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.

 

[haha1234]

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.

Like this?

 

[ehild]

Where is the force of friction?

ehild

 

[Doc Al]

Like this?

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.

 

[haha1234]

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.

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

 

[ehild]

W is vertical, it does not have horizontal component. ehild

 

[haha1234]

W is vertical, it does not have horizontal component. ehild

So is the weight giving the centripetal acceleration?

 

[ehild]

Why do you think so?

ehild

 

[Doc Al]

How to find the horizontal component of the weight?

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

I think W_b in the picture is not parallel to the centripetal acceleration,so it may not be a component of the centripetal force.

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.)

 

[haha1234]

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.)

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

 

[haruspex]

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

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.)

 

[haha1234]

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.)

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

 

[Doc Al]

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

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.

 

[Doc Al]

Is the net force in vertical direction 0N?

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

Is the net force in horizontal direction Nsinθ+fcosθ?

Yes, for one of the extremes.

 

[haha1234]

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.

N_y+W=0?

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

 

[Doc Al]

N_y+W=0?

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

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

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.

 

[haha1234]

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.

Vertical component:
N_y-W-fsin=0?

Horizontal component:
N_x+fcosθ=mv²/R?

 

[Doc Al]

Vertical component:
N_y-W-fsin=0?

Express Ny in terms of θ and N.

Horizontal component:
N_x+fcosθ=mv²/R?

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.

 

[haha1234]

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.

Nx+fcosθ=mv_max²/R
mgtanθ+μ_sNcosθ=mv_max²/R
gtanθ+μ_s(g/cosθ)cosθ=v_max²/R
v_max=[Rg(tanθ+μ_s)]^1/2
But it's not the correct answer.

 

[Doc Al]

Nx+fcosθ=mv_max²/R
mgtanθ+μ_sNcosθ=mv_max²/R

Why do you think Nx = mgtanθ?

Just call the normal force N; it is one of the unknowns. Do as I suggested: Express Nx and Ny in terms of N and θ. (Don't solve for N in terms of anything else. That comes later.)

 

[haha1234]

Why do you think Nx = mgtanθ?

Just call the normal force N; it is one of the unknowns. Do as I suggested: Express Nx and Ny in terms of N and θ. (Don't solve for N in terms of anything else. That comes later.)

Nsinθ+μ_sNcosθ=mv_max²/R
N(sinθ+μ_scosθ)=mv_max²/R
v_max=[RN(sinθ+μ_scosθ)/m]^1/2?

But why N_x not equals to mgtanθ?
I think N_y=mg,so N_x=mgtanθ.

 

[Doc Al]

Nsinθ+μ_sNcosθ=mv_max²/R

Good! That equation comes from applying Newton's 2nd law in the horizontal direction. Now get a second equation by applying Newton's 2nd law in the vertical direction.

 

[Doc Al]

But why N_x not equals to mgtanθ?
I think N_y=mg,so N_x=mgtanθ.

Well, you are wrong. Don't guess! Write the equations and solve for the unknowns. (The only one you need to solve for is the velocity.)

 

[haha1234]

Good! That equation comes from applying Newton's 2nd law in the horizontal direction. Now get a second equation by applying Newton's 2nd law in the vertical direction.

Thank you!
I can find the correct answer now!