Chapter 5: Applications of Newton's laws
Tuesday, September 17, 2013
10:00 PM
General strategy for using Newton's second law to solve problems:
1.
2.
3.
4.
5.
Draw a diagram; select a coördinate system
Identify relevant objects/agents
Sketch an interaction scheme
Draw a free-body diagram, using the coördinate system chosen in Step 1
Apply Newton's second law in each direction specified by the coördinate system chosen in Step 1
Example: Static equilibrium
Calculate the tension in the strings if
a. the labelled angles are both 30°
b. the labelled angles are both 10°
c. one labelled angle is 40° and the other labelled angle is 60°
Mass and Weight
Apparently we don't "feel" gravitational forces directly; what we do feel is contact forces. This
explains why we feel weightless when falling, and why astronauts feel weightless when they are in
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orbit around the Earth (in "free fall").
When you stand on your bathroom scale, you "feel" the normal force exerted by the scale on you;
this is what the scale also measures. You feel a similar normal force when you sit on a chair. If you
are just standing on your bathroom scale, the normal force from the scale balances your weight
(i.e., the gravitational force that the Earth exerts on you), which is why the scale reading (which
measures this normal force) equals your weight.
Make sure you understand the difference between mass and weight; your mass is a measure of
the substances that you consist of, and is independent of your location. Your weight is a measure
of the gravitational force that the Earth exerts on you; although it's proportional to your mass, it is
not the same as your mass (W = mg). If you were on the Moon, your weight would be different,
because what you feel as your weight would be the normal force that the surface of the Moon
exerts on you, which is equal to the gravitational force that the Moon exerts on you.
When you are accelerating, your apparent weight (i.e., the normal force exerted on you by the
surface you're standing on) depends on your acceleration.
Example: Apparent weight
Calculate the apparent weight of a 70 kg person in an elevator that is
a. accelerating upward at 4 m/s2.
b. moving upward at a constant speed of 5 m/s.
c. accelerating downward at 2 m/s2.
Friction
You may have noticed that it is difficult to push a refrigerator. The reason for this is that the bottom of the
refrigerator is rough, and so is the floor.
You may also have noticed that it's more difficult to move the refrigerator over a carpet than over a
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You may also have noticed that it's more difficult to move the refrigerator over a carpet than over a
smooth floor, because the carpet is rougher than the floor. Thus, the frictional force between two surfaces
depends on the surfaces. This dependence is so complicated that we don't have any good theory for
predicting the friction between two surfaces; the best we can do is just measure the frictional forces in
experiments.
Further reflection will reveal other facts about friction. You may have noticed that when you begin to push
on the refrigerator, it doesn't move. However, as you gradually increase your pushing force, eventually the
refrigerator will move. Once the refrigerator is moving, it takes less pushing force to keep it moving than
the force needed to get it moving in the first place.
Conclusion: the coefficient of static friction is greater than the coefficient of kinetic friction.
Furthermore, if you stack another heavy object on the refrigerator, it will be harder to move it. Similarly, if
someone presses down on the refrigerator, that also makes it harder to move. Thus, frictional forces
depend on the normal force acting between the two surfaces in contact.
Conclusions:
fs ≤ µsn
fk = µkn
Typically, the coefficient of kinetic friction is less than the coefficient of static friction for the same pair of
surfaces; the coefficient of rolling friction is much less than the coefficient of static friction for the same
surfaces.
µr << µk < µs
Sample coefficients of friction are found on Page 144 of your textbook. For example, for rubber on
concrete, approximate typical values are
µs = 1.00, µk = 0.80, µr = 0.02
Example: A refrigerator of mass 100 kg is resting on a horizontal floor. The coefficients of friction between
the refrigerator and the floor are µs = 1.00 and µk = 0.73. Determine
a. the force needed to get the refrigerator moving.
b. the force needed to keep the refrigerator moving at a constant speed once it has already started moving.
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Air drag
• very complicated, much like friction; we don't have good theories of air drag, so we can't predict very well
the amount of air drag that will be present in some situation; the best we can do is to experiment,
sometimes with scale models (wind tunnels, for example, are used for experimentation)
As an approximation, that is reasonably good in certain circumstances, the force of air drag (i.e., air
resistance) in newtons is about:
D = 0.3Av2
where A is the cross-sectional area of the object that is moving through the air (in square metres), and the
speed is measured in m/s. Notice the quadratic dependence on speed; this means if the speed doubles, the
air drag quadruples. This makes it very difficult indeed to get things moving through the air at high speeds.
A lot of engineering design work goes into reducing air drag by designing shapes and using materials for
objects that will reduce the coefficient "0.3" to a smaller number. A more detailed formula is
D = 0.25ρAv2
where ρ represents the density of air; a typical value for the density of air at sea level is ρ = 1.22 kg/m3.
A consequence of air drag is terminal speed. If you drop an object from rest, the gravitational force on the
object is approximately constant for the entire fall, but the air drag starts off very small (because the speed
is small) and then increases rapidly with increasing speed. At some point, if the object falls for long enough,
the air drag will grow so large that it balances the gravitational force. After this point, the net force acting
on the falling object will be zero, so the object's speed will be constant; this is called the object's terminal
speed.
Example: Terminal speed
a. Calculate the terminal speed of a human being.
b. Calculate the area of a parachute needed to reduce the terminal speed of a human being to (i) 10 km/h,
and (ii) 5 km/h.
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Block and pulleys
Example: Calculate the acceleration of the system and the inter-block forces if
a. a 10 N force acts on the left-most block towards the right.
b. a 10 N force acts on the right-most block towards the left.
Example: Calculate the tension in the string and the acceleration of each block. Assume that the strings are
massless and don't stretch, and that the pulley is massless and there is no friction at the bearing. Further
assume that the string slides on the pulley without friction (or turns the pulley without friction needing to
be considered).
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Example: Calculate the tension in the string and the acceleration of each block. Assume that the
strings are massless and don't stretch, and that the pulley is massless and there is no friction at the
bearing. Further assume that the string slides on the pulley without friction (or turns the pulley
without friction needing to be considered). Suppose that the coefficient of static friction between the
2 kg block and the horizontal surface is 0.8, and that the coefficient of kinetic friction between the
same surfaces is 0.5.
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Example: Calculate the tension in the string and the acceleration of each block. Assume that the strings
are massless and don't stretch, and that the pulley is massless and there is no friction at the bearing.
Further assume that the string slides on the pulley without friction (or turns the pulley without friction
needing to be considered). Suppose that the coefficient of static friction between the 2 kg block and the
horizontal surface is 0.8, and that the coefficient of kinetic friction between the same surfaces is 0.5.
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