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)
Thanks for the reply!
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?
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
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.
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?
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.
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?