The advisory speed for a car on a sloped bend

[Grizzly_1]

Homework Statement

At bends on motorways the road is sloped so that a car is less likely to slide out of its
lane when travelling at a high speed.

Figure 7 shows a car of mass 1200 kg travelling around a curve of radius 200 m.
The motorway is sloped at an angle of 5.0°.

Figure 8 shows the weight W and reaction force N acting on the car. The advisory
speed for the bend is chosen so that the friction force down the slope is zero.

Suggest an appropriate advisory speed for this section of the motorway

Relevant Equations

F=mv^2/r

Hello all, I am sadly stuck on the last part of a circular motion question sheet I was given for homework. I have a mark-scheme with me, but it has actually given me more questions than answers. I have attached my working, and how I arrived at my answer, and the differences it has with the mark-scheme answer. To note, the mark-scheme answer rounds to my answer, but the trigonometry they used is entirely opposite to mine. I would really appreciate some help in understanding:

Thanks to anyone who can lend me a hand on this. I assume it is just poor physics on my end, so getting through this will help keep me up to scratch on circular motion.

 

[erobz]

Homework Statement: At bends on motorways the road is sloped so that a car is less likely to slide out of its
lane when travelling at a high speed.

Figure 7 shows a car of mass 1200 kg travelling around a curve of radius 200 m.
The motorway is sloped at an angle of 5.0°.

Figure 8 shows the weight W and reaction force N acting on the car. The advisory
speed for the bend is chosen so that the friction force down the slope is zero.

Suggest an appropriate advisory speed for this section of the motorway
Relevant Equations: F=mv^2/r

Hello all, I am sadly stuck on the last part of a circular motion question sheet I was given for homework. I have a mark-scheme with me, but it has actually given me more questions than answers. I have attached my working, and how I arrived at my answer, and the differences it has with the mark-scheme answer. To note, the mark-scheme answer rounds to my answer, but the trigonometry they used is entirely opposite to mine. I would really appreciate some help in understanding:
View attachment 325957View attachment 325958View attachment 325959
Thanks to anyone who can lend me a hand on this. I assume it is just poor physics on my end, so getting through this will help keep me up to scratch on circular motion.

I get $13.10 \frac{m}{s}$, is that the correct answer?

 

[Grizzly_1]

I get $13.10 \frac{m}{s}$, is that the correct answer?

Well yes, 13.1 is the answer from the mark-scheme (see the second attachment), my problem is how you arrive at this result. You must have resolved the forces differently to me, and if you don't mind I would like to know how/why you did it this way. Thank you for your reply!

 

[erobz]

Well yes, 13.1 is the answer from the mark-scheme (see the second attachment), my problem is how you arrive at this result. You must have resolved the forces differently to me, and if you don't mind I would like to know how/why you did it this way. Thank you for your reply!

I put my coordinates parallel and perpendicular to the slope. Then in the non-inertial frame of the car did a force balance parallel to the slope. Care to try that, and see what you get?

EDIT: Oh, and please use Latex to write your equations (see the LaTeX Guide in the lower left corner of your reply). Pictures are supposed to be for diagrams. It's hard to read handwritten equations, and it makes it easier to quote any parts of your working that may contain an error.

 

[erobz]

If you don't want to go that route, then in your approach ask yourself "What direction is the radial acceleration" That's where you made the mistake (as far as I can tell).

 

[Grizzly_1]

I put my coordinates parallel and perpendicular to the slope. Then in the non-inertial frame of the car did a force balance parallel to the slope. Care to try that, and see what you get?

EDIT: Oh, and please use Latex to write your equations (see the LaTeX Guide in the lower left corner of your reply). Pictures are supposed to be for diagrams. It's hard to read handwritten equations, and it makes it easier to quote any parts of your working that may contain an error.

If you don't want to go that route, then in your approach ask yourself "What direction is the radial acceleration" That's where you made the mistake.

Thank you for both of these replies, I would think the centripetal acceleration is horizontally towards the centre. I apologise for not using LaTeX, I will get up to scratch to use it next time. Finally I am unsure of what you mean in your first reply.

 

[kuruman]

You have an expression of Newton's second law for the motion in the horizontal direction. That is $\cancel{N\cos\!\varphi=\dfrac{mv^2}{r}}$ and I agree.

How about writing an expression for Newton's second law in the vertical direction?

On edit:
The correct equation is
$N\sin\!\varphi=\dfrac{mv^2}{r}$.

 

[erobz]

Thank you for both of these replies, I would think the centripetal acceleration is horizontally towards the centre. I apologise for not using LaTeX, I will get up to scratch to use it next time. Finally I am unsure of what you mean in your first reply.

Did you resolve ALL vectors parallel and normal to the slope (the radial acceleration is a vector too)

 

[erobz]

You have an expression of Newton's second law for the motion in the horizontal direction. That is $N\cos\!\varphi=\dfrac{mv^2}{r}$ and I agree.

How about writing an expression for Newton's second law in the vertical direction?

Don't you mean $N \sin \varphi = \frac{mv^2}{r}$?

 

[berkeman]

I apologise for not using LaTeX, I will get up to scratch to use it next time.

Sending you a PM with LaTeX tips...

 

[haruspex]

I assume it is just poor physics on my end,

Your mistake is the line $W\cos(\theta)=N$. For that, you presumably resolved normal to the slope, but the centripetal acceleration has a component normal to the slope, so those two forces are not in balance.

 

[kuruman]

Don't you mean $N \sin \varphi = \frac{mv^2}{r}$?

Yes, that's what meant for the horizontal component. Thanks.