Last time, we setup a simple example to work through of a block sliding down a ramp:
Taylor does this as his example 1.1, but he takes his \( x \)-axis to be parallel to the ramp. To get used to comparing different coordinate systems, I'll instead set my \( y \)-axis to be in the direction of gravity - I'll call this the \( y' \) axis. As Taylor points out, this will make things slightly more complicated, but it's not too bad. We'll treat this as a two-dimensional problem, as indicated, which just means that we ignore the \( z \) coordinate. (This is valid because of Newton's first law: if there are no \( z \)-direction forces, then there is no interesting \( z \)-direction motion at all.)
To proceed, we'll need to split all the forces into components. Let's draw some angles on our free-body diagram:
(if where I put the \( \theta \)'s in my diagram isn't obvious to you, draw more parallel lines in and use right triangles to identify which angles are \( \theta \) and which are \( 90^\circ - \theta \).) From the diagram, the forces split into components are, using the often-convenient notation that \( \vec{F} = (F_x, F_y) \):
\[ \begin{aligned} \vec{N} = (N \sin \theta, N \cos \theta) \\ \vec{F}_f = (-F_f \cos \theta, F_f \sin \theta) \\ \vec{F}_g = (0, -mg) \end{aligned} \]
Now, we know that the normal force \( N \) is equal and opposite to the magnitude of all other forces acting perpendicular to the surface of the ramp. From the diagram, we can just read off that it is equal to a component of the gravitational force,
\[ \begin{aligned} N = mg \cos \theta. \end{aligned} \]
The second fact we know is that the magnitude of the frictional force is proportional to the normal force,
\[ \begin{aligned} F_f = \mu N = \mu mg \cos \theta. \end{aligned} \]
Plugging back in, then, we have the net forces in the \( x' \) and \( y' \) directions:
\[ \begin{aligned} \vec{F}_{\rm net} = \vec{N} + \vec{F}_f + \vec{F}_g \\ = \left(N \sin \theta - F_f \cos \theta \right) \hat{x}' + \left(N \cos \theta + F_f \sin \theta - mg \right) \hat{y}' \\ = mg \left( \cos \theta \sin \theta - \mu \cos^2 \theta \right) \hat{x}' + mg \left(\cos^2 \theta + \mu \cos \theta \sin \theta - 1 \right) \hat{y}' \end{aligned} \]
This looks a little messy, but let's keep going. Now that we have the net force, we can solve for the accelerations using \( \vec{F}_{\rm net} = m \ddot{\vec{r}} \), which gives us two equations looking at each component (and cancelling off the mass):
\[ \begin{aligned} \ddot{x}' = g \cos \theta (\sin \theta - \mu \cos \theta) \\ \ddot{y}' = -g (1 - \cos^2 \theta - \mu \sin \theta \cos \theta) \\ = -g \sin \theta (\sin \theta - \mu \cos \theta) \end{aligned} \]
where I've used a trig identity to simplify the last line; now the \( x' \) and \( y' \) equations look very similar.
The good news is that even though these still don't look so nice, everything on the right-hand side of both equations is constant, so to solve for the motion, all we have to do is integrate twice, for example:
\[ \begin{aligned} \dot{x}'(t) = v_{x,0} + \int_0^{t} dt' \ddot{x}'(t') = v_{x',0} + gt \cos \theta (\sin \theta - \mu \cos \theta) \\ x'(t) = \int_0^t dt' \dot{x}'(t') = x'_0 + v_{x',0} t + \frac{1}{2} gt^2 \cos \theta (\sin \theta - \mu \cos \theta) \end{aligned} \]
and similarly,
\[ \begin{aligned} y'(t) = y'_0 + v_{y',0} t -\frac{1}{2} gt^2 \sin \theta (\sin \theta - \mu \cos \theta). \end{aligned} \]
If we assume starting from rest, then \( v_{x',0}, v_{y',0} \) are set to zero; we'll do that from here forward, although our solution can handle a starting speed as well. (One tricky little detail: we've assumed that the block stays on the surface of the ramp for the whole motion. This restricts the direction of the initial velocity, which means we can't just pick any arbitrary \( v_{x',0} \) and \( v_{y',0} \).)
Let's check some limits! (I leave it to you to verify that the units all make sense.) If \( \theta \rightarrow 90^\circ \), then the ramp becomes completely vertical, so the motion should only be in the \( y' \) direction, and it should just be freefall. At \( 90^\circ \) we have \( \cos \theta = 0 \) and \( \sin \theta = 1 \), so
\[ \begin{aligned} x'(t) \xrightarrow{\theta = 90^\circ} x_0' + \frac{1}{2} gt^2 (0) (1 - 0\mu) = x_0' \\ y'(t) \xrightarrow{\theta = 90^\circ} y_0' - \frac{1}{2} gt^2 (1) (1 - 0\mu) = y_0' - \frac{1}{2} gt^2 \end{aligned} \]
so indeed, we have freefall in the \( y' \) direction and no motion in the \( x' \) direction.
How about the opposite limit, \( \theta \rightarrow 0^\circ \)? Since the motion is entirely caused by gravity, what we expect to find is that as the ramp becomes completely flat, all motion stops. Plugging in again, we find this time that
\[ \begin{aligned} x'(t) \xrightarrow{\theta = 0^\circ} x_0' + \frac{1}{2} gt^2 (1) (0 - 1\mu) = x_0' - \frac{1}{2} \mu gt^2 \\ y'(t) \xrightarrow{\theta = 0^\circ} y_0' - \frac{1}{2} gt^2 (0) (0 - 1\mu) = y_0' \end{aligned} \]
The \( y'(t) \) result is fine, but our block will apparently start sliding backwards in the \( x' \) direction if the ramp is laid flat! What went wrong with our calculation?! (Think about it yourself, before you continue reading...)
In fact, our calculation above is just fine. The problem that was revealed when \( \theta \rightarrow 0^\circ \) has to do with our starting assumptions. In particular, we assumed a single coefficient of friction \( \mu \). Since our block is in motion, this must be the coefficient of kinetic friction, \( \mu_k \). However, we know that there is also a coefficent of static friction \( \mu_s \) that must be overcome if the block is starting at rest,
\[ \begin{aligned} F_{f,s} \leq \mu_s mg \cos \theta, \end{aligned} \]
which will keep the block from moving (i.e. give zero accelerations) if the other forces are not large enough. We can take the \( x' \)-direction net force and solve:
\[ \begin{aligned} F_{\rm net, x'} = N \sin \theta - F_f \cos \theta = 0 \\ mg \cos \theta \sin \theta = F_f \cos \theta \\ mg \sin \theta \leq \mu_s mg \cos \theta \\ \tan \theta \leq \mu_s \end{aligned} \]
So the full picture is: if the angle \( \theta \) is small enough that \( \tan \theta \leq \mu_s \), then the block just won't move at all. Since generally \( \mu_k \leq \mu_s \), this means that \( \tan \theta \geq \mu_k \) whenever static friction is overcome; this means that the offending term which changed the sign of our \( x' \)-direction motion,
\[ \begin{aligned} (\sin \theta - \mu_k \cos \theta) \geq \sin \theta - \tan \theta \cos \theta \geq 0 \end{aligned} \]
and we don't have any problems with backwards-moving blocks anymore.
One last thing we should do is compare our answer to Taylor's answer, which means we need to change coordinates.
Clicker Question
From the diagram above, what is \( x(t) \) in terms of \( x'(t) \) and \( y'(t) \)? (Assume the origin of the two coordinate systems is the same.)
A. \( x(t) = x'(t) \sin \theta \)
B. \( x(t) = x'(t) \cos \theta \)
C. \( x(t) = x'(t) \sin \theta - y'(t) \cos \theta \)
D. \( x(t) = x'(t) \cos \theta - y'(t) \sin \theta \)
E. It depends on the time \( t \).
Answer: D
This is a nice exercise, because it really drives home the point I started this whole discussion with: to compare our calculation to Taylor, we have to figure out how to put them into the same coordinate system!
First of all, E can't be correct; because the origin of the coordinate systems is the same fixed point, the coordinate axes never change with \( t \).
To solve this, we should first add the angle \( \theta \) on to our coordinate diagram:
For a point along the \( x \) axis, we can certainly see that it will have both an \( x' \) component (positive) and a \( y' \) component (negative), narrowing our options down to C or D.
To distinguish the two remaining options, we can either draw right triangles, or think in terms of dot products: the angle \( \theta \) as drawn is between \( \hat{x} \) and \( \hat{x}' \), so we read off
\[ \begin{aligned} \hat{x} \cdot \hat{x}' = \cos \theta, \hat{x} \cdot \hat{y}' = \cos (\frac{\pi}{2} + \theta) = -\sin \theta. \end{aligned} \]
Then for any position vector \( \vec{r} \), we can decompose in either coordinate system,
\[ \begin{aligned} \vec{r} = x \hat{x} + y\hat{y} = x' \hat{x}' + y' \hat{y}' \end{aligned} \]
and \( \hat{x} \cdot \vec{r} \) will give us \( x(t) \), specifically option D.
Putting our solutions together, then, we have (starting from rest and the origin as we were told)
\[ \begin{aligned} x(t) = \cos \theta \left(\frac{1}{2} gt^2 \cos \theta (\sin \theta - \mu \cos \theta)\right) - \sin \theta \left( -\frac{1}{2} gt^2 \sin \theta (\sin \theta - \mu \cos \theta) \right) \\ = \frac{1}{2} gt^2 (\sin \theta - \mu \cos \theta) \end{aligned} \]
and if we open up the textbook, we'll find that this matches Taylor's solution exactly.
Reference frames and mechanics
Whenever we do classical mechanics, we have to specify a reference frame. Reference frames are choices of coordinate systems, but because we're dealing with both time AND space, they have an extra complication as compared to fixed coordinates: two frames can be moving relative to one another. Following Taylor, we'll use a script \( \mathcal{S} \) to denote a reference frame. \( \mathcal{S} \) includes :
- A coordinate system,
- An origin \( \mathcal{O} \), and finally
- Some specification of how \( \mathcal{O} \) is or isn't moving, relative to some physical object or other point of reference.
Our chosen coordinates \( (x', y') \) for the ramp problem are an example of a reference frame \( \mathcal{S}' \); Taylor's rotated coordinates \( (x,y) \) are a different reference frame \( \mathcal{S} \). Both frames have the same origin, and both are fixed with respect to the ramp, which means there's also no relative motion between \( \mathcal{S} \) and \( \mathcal{S}' \). However, we could come up with a third frame \( \mathcal{S}'' \) with coordinates \( (x'', y'') \), defined by the following relations:
\[ \begin{aligned} x''(t) = x'(t) - ut \\ y''(t) = y'(t) \end{aligned} \]
In other words, the origin of \( \mathcal{S}'' \) is moving horizontally to the right with constant speed \( u \).
Since we already solved for the motion in \( \mathcal{S}' \), we can just use the coordinate change to find the motion in \( \mathcal{S}'' \):
\[ \begin{aligned} x''(t) = x''_0 + (v_{x',0} - u) t + \frac{1}{2} gt^2 \cos \theta (\sin \theta - \mu \cos \theta) \\ y'(t) = y''_0 + v_{y',0} t -\frac{1}{2} gt^2 \sin \theta (\sin \theta - \mu \cos \theta). \end{aligned} \]
Notice that if our block starts at rest with respect to the ramp, it will appear to be moving to the left with initial speed \( u \) in our moving coordinates.
Just using a change of frame on our solution is always valid; if we have solved the equations of motion in one reference frame, we know the answer in any other frame. However, we have to be careful if we start in a given reference frame and try to apply Newton's laws. If you try it in the \( \mathcal{S}'' \) frame, the forces will all be exactly the same, and you'll find the same answer that we got above. However, consider a different example:
We drop a ball inside an elevator, from initial height \( y_0 \). From the point of view of frame \( \mathcal{S} \), fixed with respect to the ground, the elevator is moving straight down with constant acceleration \( a \). As pictured, we can define a second frame of reference \( \mathcal{S}' \) which is fixed to the elevator (so the point of view of an experimenter riding the elevator, basically.)