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# Another momentum question Watch

1. I have another momentum question, 10a attached

I've also attached a modified version of the diagram displaying what I think I need to do

I simply do not understand why mgcos20 doesn't give me the right answer

thanks for any help!
Attached Images

2. (Original post by ihateocr12345)
I have another momentum question, 10a attached

I've also attached a modified version of the diagram displaying what I think I need to do

I simply do not understand why mgcos20 doesn't give me the right answer

thanks for any help!
This is what the mark scheme says, I do understand their method, however I don't understand why I can't do it the way I suggested above
3. (Original post by ihateocr12345)
This is what the mark scheme says, I do understand their method, however I don't understand why I can't do it the way I suggested above
R is going to be greater than the weight... the car is in vertical equilibrium so the vertical component of R=mg

cos (θ) only gives values from -1 to +1, the maximum value of mg cos (θ) is equal to mg (and that's when θ = 0)
4. (Original post by Joinedup)
R is going to be greater than the weight... the car is in vertical equilibrium so the vertical component of R=mg

cos (θ) only gives values from -1 to +1, the maximum value of mg cos (θ) is equal to mg (and that's when θ = 0)

https://www.youtube.com/watch?v=dA4BvYdw7Xg - According to this guy, and many other people, it's done the way I have done. And to be honest, I think it may be an error in the mark scheme. Surely it would make sense for R = mg when θ = 0, and for R < mg when θ > 0. When θ is gradually increased to 90, the reaction force surely gradually decreases to 0, and equals 0 when θ = 90?
5. (Original post by ihateocr12345)
I have another momentum question, 10a attached

I've also attached a modified version of the diagram displaying what I think I need to do

I simply do not understand why mgcos20 doesn't give me the right answer

thanks for any help!

I formed the triangle differently to you (check diagram above).

Angle at the bottom = 20 degrees, to find the force acting perpendicular to the slope, I used trig:

Hypotenuse: 8044.2 N

CAH
cos(20) = a / 8044.2
a = 8044.2 cos(20)
a = 7559 N

I think the answer they've given is wrong if you got the same answer as me?
6. Have you tried sin theta?
7. I had literally the same problem yesterday
Let me explain it to you, the car is travelling around the bank, this means that it has a horizontal acceleration (centripetal acceleration), correct? The normal to the bank, i.e. the direction of R, is NOT perpendicular to the centripetal acceleration, this means that the centripetal acceleration has a component in the direction perpendicular to the bank. So,
R - MGcostheta does not equal zero, it equals the component of the centripetal force perpendicular to the bank which is NOT zero. This is why you must solve in the vertical direction. In the vertical direction, there is no force as the vertical direction is perpendicular to the centripetal force. Also, if there was a force in the vertical direction the car would fly
8. (Original post by oShahpo)
I had literally the same problem yesterday
Let me explain it to you, the car is travelling around the bank, this means that it has a horizontal acceleration (centripetal acceleration), correct? The normal to the bank, i.e. the direction of R, is NOT perpendicular to the centripetal acceleration, this means that the centripetal acceleration has a component in the direction perpendicular to the bank. So,
R - MGcostheta does not equal zero, it equals the component of the centripetal force perpendicular to the bank which is NOT zero. This is why you must solve in the vertical direction. In the vertical direction, there is no force as the vertical direction is perpendicular to the centripetal force. Also, if there was a force in the vertical direction the car would fly
Reaction force is equal to mg
So resultant vertical force is zero.
9. (Original post by Kyx)
Reaction force is equal to mg
So resultant vertical force is zero.
Yea but reaction forces does not equal mgcostheta. R > Mg costheta.
10. try tan theta

reaction force of mg / reaction force of centripietal = grtantheta ?
11. (Original post by oShahpo)
I had literally the same problem yesterday
Let me explain it to you, the car is travelling around the bank, this means that it has a horizontal acceleration (centripetal acceleration), correct? The normal to the bank, i.e. the direction of R, is NOT perpendicular to the centripetal acceleration, this means that the centripetal acceleration has a component in the direction perpendicular to the bank. So,
R - MGcostheta does not equal zero, it equals the component of the centripetal force perpendicular to the bank which is NOT zero. This is why you must solve in the vertical direction. In the vertical direction, there is no force as the vertical direction is perpendicular to the centripetal force. Also, if there was a force in the vertical direction the car would fly
Thanks for the reply and explanation! still really confused

How can centripetal acceleration have a component perpendicular to the bank? I thought it acts perpendicularly to the direction of motion, which is towards the centre of the circle?
12. (Original post by derpz)

I formed the triangle differently to you (check diagram above).

Angle at the bottom = 20 degrees, to find the force acting perpendicular to the slope, I used trig:

Hypotenuse: 8044.2 N

CAH
cos(20) = a / 8044.2
a = 8044.2 cos(20)
a = 7559 N

I think the answer they've given is wrong if you got the same answer as me?

And this is the answer I got
13. (Original post by ihateocr12345)
Thanks for the reply and explanation! still really confused

How can centripetal acceleration have a component perpendicular to the bank? I thought it acts perpendicularly to the direction of motion, which is towards the centre of the circle?
Yea it does. Look, here's a general a rule, if you have a force or any vector in general, it has components in ALL directions not perpendicular to it.
For example, you can always find the component of gravity perpendicular to the surface, right? Now if the surface was perfectly vertical, gravity can't have a component perpendicular to the surface, right?

Now the centripetal force is horizontal towards the centre. The bank is slightly tilted, so that the horizontal is not perpendicular to the bank's perpendicular. This means that there is a component of the centripetal force perpendicular to the bank. Thus, R- MgCostheta does not equal zero, it equals that component. The component is in fact equal to Fcentripetal * Sintheta.
So R- MgCostheta = F Centripetal * sin Theta.
14. (Original post by oShahpo)
Yea it does. Look, here's a general a rule, if you have a force or any vector in general, it has components in ALL directions not perpendicular to it.
For example, you can always find the component of gravity perpendicular to the surface, right? Now if the surface was perfectly vertical, gravity can't have a component perpendicular to the surface, right?

Now the centripetal force is horizontal towards the centre. The bank is slightly tilted, so that the horizontal is not perpendicular to the bank's perpendicular. This means that there is a component of the centripetal force perpendicular to the bank. Thus, R- MgCostheta does not equal zero, it equals that component. The component is in fact equal to Fcentripetal * Sintheta.
So R- MgCostheta = F Centripetal * sin Theta.
i think i understand now, however were does "F Centripetal * sin Theta" come from?
15. (Original post by ihateocr12345)
i think i understand now, however were does "F Centripetal * sin Theta" come from?
Breaking up the horizontal force (centripetal) into a direction perpendicular to the bank and parallel to the bank. A bit of geometry will help.

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