4.3 Newton's second law by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.
vCalc Companion Formulas | |
vCalc Formulary | 4.3 Newton's second law |
`A=F/M` | Acceleration [Force/Mass] |
`A=(F_t-F_g)/M` | Acceleration [`F_t-F_g`/M] |
`F_"total"`=`mDeltav`/`Deltat` | Force |
|`F_W`|=`mg` | Force [`F_w`] |
`F_(drag)= 1/2*(rho*A*Cd*v^2)` | Force of Drag |
`v_(terminal)`= `sqrt((mg)/(crhoA))` | Velocity [`V_"terminal"`] |
What about cases where the total force on an object is not zero, so that Newton's first law doesn't apply? The object will have an acceleration. The way we've defined positive and negative signs of force and acceleration guarantees that positive forces produce positive accelerations, and likewise for negative values. How much acceleration will it have? It will clearly depend on both the object's mass and on the amount of force.
Experiments with any particular object show that its acceleration is directly proportional to the total force applied to it. This may seem wrong, since we know of many cases where small amounts of force fail to move an object at all, and larger forces get it going. This apparent failure of proportionality actually results from forgetting that there is a frictional force in addition to the force we apply to move the object. The object's acceleration is exactly proportional to the total force on it, not to any individual force on it. In the absence of friction, even a very tiny force can slowly change the velocity of a very massive object.
Experiments (e.g., the one described in example 11 on p. 140) also show that the acceleration is inversely proportional to the object's mass, and combining these two proportionalities gives the following way of predicting the acceleration of any object:
e / Example 6
`a = F_T / m`, where m is an object's mass
`F_"total" (F_T)` is the sum of the forces acting on it, and
`a` is the acceleration of the object's center of mass.
We are presently restricted to the case where the forces of interest are parallel to the direction of motion.
We have already encountered the SI unit of force, which is the newton (N). It is designed so that the units in Newton's second law all work out if we use SI units: m/`s^2` for acceleration and kg (not grams!) for mass.
f / A coin slides across a table. Even for motion in one dimension, some of the forces may not lie along the line of the motion.
? The Falcon 9 launch vehicle, built and operated by the private company SpaceX, has mass m=5.1×`10^5` kg. At launch, it has two forces acting on it: an upward thrust `F_t`=5.9×`10^6` N and a downward gravitational force of `F_g`=5.0×`10^6` N. Find its acceleration.
? Let's choose our coordinate system such that positive is up. Then the downward force of gravity is considered negative. Using Newton's second law,
=`((5.9× 10^6 N)-(5.0× 10^6N))/(5.1× 10^5 kg)`
=1.6 m/`s^2`,
where as noted above, units of N/kg (newtons per kilogram) are the same as m/`s^2`.
g / A simple double-pan balance works by comparing the weight forces exerted by the earth on the contents of the two pans. Since the two pans are at almost the same location on the earth's surface, the value of `g` is essentially the same for each one, and equality of weight therefore also implies equality of mass.
? A VW bus with a mass of 2000 kg accelerates from 0 to 25 m/s (freeway speed) in 34 s. Assuming the acceleration is constant, what is the total force on the bus?
? We solve Newton's second law for `F_"total"`=`ma`, and substitute `Deltav`/`Deltat` for `a`, giving
`F_"Total"`=`mDeltav`/`Deltat`
=(2000 kg)(25 m/s?0 m/s)/(34 s)
=1.5 kN.
h / Example 9.
x (m) | t (s) |
10 | 1.84 |
20 | 2.86 |
30 | 3.80 |
40 | 4.67 |
50 | 5.53 |
60 | 6.38 |
70 | 7.23 |
80 | 8.10 |
90 | 8.96 |
100 | 9.83 Discussion question D. |
As with the first law, the second law can be easily generalized to include a much larger class of interesting situations:
Suppose an object is being acted on by two sets of forces, one set lying parallel to the object's initial direction of motion and another set acting along a perpendicular line. If the forces perpendicular to the initial direction of motion cancel out, then the object accelerates along its original line of motion according to a=F_{∥}/m, where F_{∥} is the sum of the forces parallel to the line.
Suppose a coin is sliding to the right across a table, f, and let's choose a positive x axis that points to the right. The coin's velocity is positive, and we expect based on experience that it will slow down, i.e., its acceleration should be negative.
Although the coin's motion is purely horizontal, it feels both vertical and horizontal forces. The Earth exerts a downward gravitational force `F_2` on it, and the table makes an upward force `F_3` that prevents the coin from sinking into the wood. In fact, without these vertical forces the horizontal frictional force wouldn't exist: surfaces don't exert friction against one another unless they are being pressed together.
Although `F_2`and `F_3`contribute to the physics, they do so only indirectly. The only thing that directly relates to the acceleration along the horizontal direction is the horizontal force: `a=F_1`/m.
Mass is different from weight, but they're related. An apple's mass tells us how hard it is to change its motion. Its weight measures the strength of the gravitational attraction between the apple and the planet earth. The apple's weight is less on the moon, but its mass is the same. Astronauts assembling the International Space Station in zero gravity couldn't just pitch massive modules back and forth with their bare hands; the modules were weightless, but not massless.
We have already seen the experimental evidence that when weight (the force of the earth's gravity) is the only force acting on an object, its acceleration equals the constant `g`, and `g` depends on where you are on the surface of the earth, but not on the mass of the object. Applying Newton's second law then allows us to calculate the magnitude of the gravitational force on any object in terms of its mass:
(The equation only gives the magnitude, i.e. the absolute value, of `F_W`, because we're defining `g` as a positive number, so it equals the absolute value of a falling object's acceleration.)
? Solved problem: Decelerating a car — problem 7
? Figure `h` shows masses of one and two kilograms hung from a spring scale, which measures force in units of newtons. Explain the readings.
? Let's start with the single kilogram. It's not accelerating, so evidently the total force on it is zero: the spring scale's upward force on it is canceling out the earth's downward gravitational force. The spring scale tells us how much force it is being obliged to supply, but since the two forces are equal in strength, the spring scale's reading can also be interpreted as measuring the strength of the gravitational force, i.e., the weight of the one-kilogram mass. The weight of a one-kilogram mass should be
=(1.0 kg)(9.8 m/`s^2`)=9.8 N,
and that's indeed the reading on the spring scale.
Similarly for the two-kilogram mass, we have
|`F_W`|=`mg`.
=(2.0 kg)(9.8 m/s2)=19.6 N.
? Experiments show that the force of air friction on a falling object such as a skydiver or a feather can be approximated fairly well with the equation |F_{air}|=c⋅ρ⋅A⋅v^{2}, where c is a constant, ρ is the density of the air, A is the cross-sectional area of the object as seen from below, and `v` is the object's velocity. Predict the object's terminal velocity, i.e., the final velocity it reaches after a long time.
? As the object accelerates, its greater v causes the upward force of the air to increase until finally the gravitational force and the force of air friction cancel out, after which the object continues at constant velocity. We choose a coordinate system in which positive is up, so that the gravitational force is negative and the force of air friction is positive. We want to find the velocity at which
`F_"air"`+`F_W`=`0,i.e.`,
`crhoAv^2`?mg=0.
Solving for `v` gives
`v_"terminal"`=`sqrt((mg)/(crhoA))` .
self-check:
It is important to get into the habit of interpreting equations. This may be difficult at first, but eventually you will get used to this kind of reasoning.
(1) Interpret the equation `v_"terminal"`= `sqrt((mg)/(crhoA))` . in the case of `rho`=0.
(2) How would the terminal velocity of a 4-cm steel ball compare to that of a 1-cm ball?
(3) In addition to teasing out the mathematical meaning of an equation, we also have to be able to place it in its physical context. How generally important is this equation?
(answer in the back of the PDF version of the book)
Because the force `mg` of gravity on an object of mass `m` is proportional to `m`, the acceleration predicted by Newton's second law is `"a=F/m=mg/m=g"` , in which the mass cancels out. It is therefore an ironclad prediction of Newton's laws of motion that free fall is universal: in the absence of other forces such as air resistance, heavier objects do not fall with a greater acceleration than lighter ones. The experiment by Galileo at the Leaning Tower of Pisa (p. 96) is therefore consistent with Newton's second law. Since Galileo's time, experimental methods have had several centuries in which to improve, and the second law has been subjected to similar tests with exponentially improving precision. For such an experiment in 1993,^{2} physicists at the University of Pisa (!) built a metal disk out of copper and tungsten semicircles joined together at their flat edges. They evacuated the air from a vertical shaft and dropped the disk down it 142 times, using lasers to measure any tiny rotation that would result if the accelerations of the copper and tungsten were very slightly different. The results were statistically consistent with zero rotation, and put an upper limit of 1×10^{-9} on the fractional difference in acceleration |g_{copper}-g_{tungsten}|/g.
? Show that the Newton can be reexpressed in terms of the three basic mks units as the combination kg?m/`s^2`kg?m/`s^2`.
? What is wrong with the following statements?
(1) “g is the force of gravity.”
(2) “Mass is a measure of how much space something takes up.”
? Criticize the following incorrect statement:
“If an object is at rest and the total force on it is zero, it stays at rest. There can also be cases where an object is moving and keeps on moving without having any total force on it, but that can only happen when there's no friction, like in outer space.”
? Table i gives laser timing data for Ben Johnson's 100 m dash at the 1987 World Championship in Rome. (His world record was later revoked because he tested positive for steroids.) How does the total force on him change over the duration of the race?
4.3 Newton's second law by Benjamin Crowell, Light and Matter licensed under the Creative Commons Attribution-ShareAlike license.