Resolving forces on inclined circular motion

When resolving forces vertically and horizontally in a problem where a car is going around a banked bend, why do you consider the components of the normal force? Shouldn’t the normal force just cancel with the perpendicular component of the weight?

• Yes, exactly. Thanks. – Max Allen Oct 11 at 13:15

I am going to assume no friction on the incline, since it can be added to what is said below, and does not add anything to the main ideas here.

I have included a free body diagram of the car below:

The thing to notice is the coordinate system we have chosen to use. The most useful orientation of coordinates is typically such that one axis is in the same direction as the acceleration of the object. The acceleration in this case is not down the ramp like in other inclined plane problems. Since we are undergoing circular motion, the acceleration is instead towards the center of the circle (to the left in the diagram). We therefore break the normal force into components along the acceleration ($$N_x$$) and perpendicular to the acceleration ($$N_y$$)$$^*$$.

The usefulness is best seen by applying Newton's second law for each direction, knowing that the acceleration is only along the x-axis: $$\sum F_x=-N_x=ma$$ $$\sum F_y=N_y-w=0$$

Here we see that just the vertical component of the normal force ends up balancing out the weight, and that the horizontal component of the normal force is responsible for the centripetal acceleration. This is also a pretty simple system of equations we can use to solve for the acceleration if we express the normal force components in terms of $$N$$ and trig functions.

In any case, if we chose a coordinate system not oriented along the acceleration, then you would instead have to break the acceleration itself into components, which is typically more work for you if you are trying to find the overall acceleration.

Let's explore this and use a coordinate system such that the x-axis is parallel to the ramp, like in our typical inclined plane problems. I have included a new diagram below of this:

I have included the acceleration vectors this time as well. Let's apply N2L to this new coordinate system. $$\sum F_x=-w_x=ma_x$$ $$\sum F_y=N-w_y=ma_y$$

Notice how now we have two equations dealing with the acceleration instead of one. This is not incorrect, but we just have to do a little bit more work if we want to determine the magnitude of the acceleration here. We have two vectors broken up into components here (acceleration and weight) instead of just the one in the previous case (the normal force).

We can also see here that the normal force is not equal and opposite to $$w_y$$ like we usually see in inclined plane problems. This is because now $$a_y\neq0$$. The acceleration has components both along and perpendicular to the incline.

$$^*$$Compare this to what you are probably more used to with your typical inclined plane problem (like a block sliding down a ramp). In this case the acceleration is along the ramp, so we break up the weight into different components instead, one along the acceleration (along the incline) and one perpendicular to the acceleration.

• Nice answer, might be worthy to note that you are assuming zero sideway friction from the tires (which could be a fair assumption). – Orbit Oct 11 at 20:30
• @Orbit. Ah yes I was going to do this while typing it out, then I completely forgot to. Thanks! – Aaron Stevens Oct 11 at 20:47