Week 1: Newton’s Laws |
69 |
For your homework you will do a more general case of this, one where the cannonball (or golf ball, or arrow, or whatever) is fired o of the top of a cli of height H. The solution will proceed identically except that the initial and final conditions may be di erent. In general, to find the time and range in this case one will have to solve a quadratic equation using the quadratic formula (instead of simple factorization) so if you haven’t reviewed or remembered the quadratic formula before now in the course, please do so right away.
1.8.2: The Inclined Plane
The inclined plane is another archetypical problem for motion in two dimensions. It has many variants. We’ll start with the simplest one, one that illustrates a new force, the normal force. Recall from above that the normal force is whatever magnitude it needs to be to prevent an object from moving in to a solid surface, and is always perpendicular (normal) to that surface in direction.
In addition, this problem beautifully illustrates the reason one selects coordinates aligned with the total force when that direction is consistent throughout a problem, if at all possible.
Example 1.8.2: The Inclined Plane
y |
|
θ |
N |
m |
|
|
mg |
H |
L |
|
|
|
θ |
|
x |
Figure 10: This is the naive/wrong coordinate system to use for the inclined plane problem. The problem can be solved in this coordinate frame, but the solution (as you can see) would be quite di cult.
A block m rests on a plane inclined at an angle of θ with respect to the horizontal. There is no friction (yet), but the plane exerts a normal force on the block that keeps it from falling straight down. At time t = 0 it is released (at a height H = L sin(θ) above the ground), and we might then be asked any of the “usual” questions – how long does it take to reach the ground, how fast is it going when it gets there and so on.
The motion we expect is for the block to slide down the incline, and for us to be able to solve the problem easily we have to use our intuition and ability to visualize this motion to select the best coordinate frame.
Let’s start by doing the problem foolishly. Note well that in principle we actually can solve the problem set up this way, so it isn’t really wrong, but in practice while I can solve it in this frame (having taught this course for 30 years and being pretty good at things like trig and calculus) it is somewhat less likely that you will have much luck if you haven’t even used trig or taken a derivative
70 |
Week 1: Newton’s Laws |
for three or four years. Kids, Don’t Try This at Home46...
In figure 10, I’ve drawn a coordinate frame that is lined up with gravity. However, gravity is not the only force acting any more. We expect the block to slide down the incline, not move straight down. We expect that the normal force will exert any force needed such that this is so. Let’s see what happens when we try to decompose these forces in terms of our coordinate system.
~
We start by finding the components of N , the vector normal force, in our coordinate frame:
Nx |
= |
N sin(θ) |
(104) |
Ny |
= |
N cos(θ) |
(105) |
~
where N = |N | is the (unknown) magnitude of the normal force.
We then add up the total forces in each direction and write Newton’s Second Law for each direction’s total force :
Fx |
= |
N sin(θ) = max |
(106) |
Fy |
= |
N cos(θ) − mg = may |
(107) |
Finally, we write our equations of motion for each direction:
ax |
= |
N sin(θ) |
(108) |
|||
|
|
|
||||
|
m |
|||||
|
|
|
|
|||
ay |
= |
|
N cos(θ) − mg |
(109) |
||
m |
||||||
|
|
|
|
|||
Unfortunately, we cannot solve these two equations as written yet. That is because we do not know the value of N ; it is in fact something we need to solve for! To solve them we need to add a condition on the solution, expressed as an equation. The condition we need to add is that the motion is down the incline, that is, at all times:
|
|
L cos(θ) − x(t) = tan(θ) |
(110) |
|||||||
|
|
|
|
|
y(t) |
|
|
|
||
must be true as a constraint47. That means that: |
|
|
|
|||||||
|
|
y(t) |
= |
(L cos(θ) − x(t)) tan(θ) |
|
|||||
|
dy(t) |
= |
− |
dx(t) |
tan(θ) |
|
||||
|
|
dt |
|
dt |
|
|
||||
d2y(t) |
|
|
d2x(t) |
|
||||||
|
|
|
|
= |
− |
|
|
tan(θ) |
|
|
|
|
dt2 |
dt2 |
|
|
|||||
|
|
ay |
= |
−ax tan(θ) |
(111) |
|||||
where we used the fact that the time derivative of L cos(θ) is zero! We can use this relation to eliminate (say) ay from the equations above, solve for ax, then backsubstitute to find ay . Both are constant acceleration problems and hence we can easily enough solve them. But yuk! The solutions we get will be so very complicated (at least compared to choosing a better frame), with both x and y varying nontrivially with time.
Now let’s see what happens when we choose the right (or at least a “good”) coordinate frame according to the prescription given. Such a frame is drawn in 11:
As before, we can decompose the forces in this coordinate system, but now we need to find the
~
components of the gravitational force as N = N yˆ is easy! Furthermore, we know that ay = 0 and
46Or rather, by all means give it a try, especially after reviewing my solution.
47Note that the tangent involves the horizontal distance of the block from the lower apex of the inclined plane,
x′ = L cos(θ) − x where x is measured, of course, from the origin.