5.3 Analysis of forces by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.
Newton's first and second laws deal with the total of all the forces exerted on a specific object, so it is very important to be able to figure out what forces there are. Once you have focused your attention on one object and listed the forces on it, it is also helpful to describe all the corresponding forces that must exist according to Newton's third law. We refer to this as “analyzing the forces” in which the object participates.
A barge is being pulled to the right along a canal by teams of horses on the shores. Analyze all the forces in which the barge participates.
Here I've used the word “floating” force as an example of a sensible invented term for a type of force not classified on the tree on p. 160. A more formal technical term would be “hydrostatic force.”
Note how the pairs of forces are all structured as “`A`'s force on `B`, `B`'s force on `A`”: ropes on barge and barge on ropes; water on barge and barge on water. Because all the forces in the left column are forces acting on the barge, all the forces in the right column are forces being exerted by the barge, which is why each entry in the column begins with “barge.”
Often you may be unsure whether you have forgotten one of the forces. Here are three strategies for checking your list:
See what physical result would come from the forces you've found so far. Suppose, for instance, that you'd forgotten the “floating” force on the barge in the example above. Looking at the forces you'd found, you would have found that there was a downward gravitational force on the barge which was not canceled by any upward force. The barge isn't supposed to sink, so you know you need to find a fourth, upward force.
Another technique for finding missing forces is simply to go through the list of all the common types of forces and see if any of them apply.
Make a drawing of the object, and draw a dashed boundary line around it that separates it from its environment. Look for points on the boundary where other objects come in contact with your object. This strategy guarantees that you'll find every contact force that acts on the object, although it won't help you to find non-contact forces.
Example 7: Fifi
`=>` Fifi is an industrial espionage dog who loves doing her job and looks great doing it. Shel leaps through a window and lands at initial horizontal speed `v_o` on a conveyor belt which is itself moving at the greater speed `v_b`. Unfortunately the coefficient of kinetic friction `Deltat`, during which the effect on her coiffure is un désastre. Find `Deltat`.
`=>` We analyze the forces:
Checking the analysis of the forces as described on p. 168:
(1) The physical result makes sense. The left-hand column consists of forces ???. We're describing the time when she's moving horizontally on the belt, so it makes sense that we have two vertical forces that could cancel. The rightward force is what will accelerate her until her speed matches that of the belt.
(2) We've included every relevant type of force from the tree on p. 160.
(3) We've included forces from the belt, which is the only object in contact with Fifi.
The purpose of the analysis is to let us set up equations containing enough information to solve the problem. Using the generalization of Newton's second law given on p. 137, we use the horizontal force to determine the horizontal acceleration, and separately require the vertical forces to cancel out.
Let positive `x` be to the right. Newton's second law gives
(`rightarrow`) [Private Equation]
Although it's the horizontal motion we care about, the only way to find `F_k`
is via the relation `F_k=mu_k*F_N`, and the only way to find `F_N` is from the `downarrow uparrow` forces. The two vertical forces must cancel, which means they have to be of equal strength:
`(downarrow uparrow) F_N - mg=0`.
Using the constant-acceleration equation `a= Deltav "/" Deltat`, we have
`Delta t =(v_b - v_o)/(mu_k*m*g"/"m)`
`Delta t = (v_b - v_0)/(mu_k*g)`.
The units check out:
`s=(m"/"s)/(m"/"s^2)`,
where `mu_k` is omitted as a factor because it's unitless.
We should also check that the dependence on the variables makes sense. If Fifi puts on her rubber ninja booties, increasing `mu_k`, then dividing by a larger number gives a smaller result for `Deltat`; this makes sense physically, because the greater friction will cause her to come up to the belt's speed more quickly. The dependence on `g` is similar; more gravity would press her harder against the belt, improving her traction. Increasing `v_b` increases `Deltat`, which makes sense because it will take her longer to get up to a bigger speed. Since `v_o` is subtracted, the dependence of `Deltat` on it is the other way around, and that makes sense too, because if she can land with a greater speed, she has less speeding up left to do.
Example 8: Forces don't have to be in pairs or at right angles
In figure u, the three horses are arranged symmetrically at 120 degree intervals, and are all pulling on the central knot. Let's say the knot is at rest and at least momentarily in equilibrium. The analysis of forces on the knot is as follows.
This analysis also demonstrates that it's all right to leave out details if they aren't of interest and we don't intend to include them in our model. We called the forces normal forces, but we can't actually tell whether they are normal forces or frictional forces. They are probably some combination of those, but we don't include such details in this model, since aren't interested in describing the internal physics of the knot. This is an example of a more general fact about science, which is that science doesn't describe reality. It describes simplified models of reality, because reality is always too complex to model exactly.
A In the example of the barge going down the canal, I referred to a “floating” or “hydrostatic” force that keeps the boat from sinking. If you were adding a new branch on the force-classification tree to represent this force, where would it go?
B The earth's gravitational force on you, i.e., your weight, is always equal to `mg`, where `m` is your mass. So why can you get a shovel to go deeper into the ground by jumping onto it? Just because you're jumping, that doesn't mean your mass or weight is any greater, does it?
5.3 Analysis of forces by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.