Forces by abstraks

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									The following is a tip sheet that should help highlight some useful strategies and common pitfalls encountered
by students studying Newtonian mechanics. It should not be used as a substitute for the textbook, lecture,
your notes, or anything else.

General physics problem solving tips
If you do not know when a formula is useful (correct and pertinent), then the formula is
useless. Becoming a good physicist does not come from memorizing a bunch of formulas
and practicing math. Strive to understand which formulas are universally valid (ex: Newton’s
laws) or definitions and which formulas only apply in special cases (ex: N = mg). Also, strive
to understand when certain problem solving methods are more useful than others.

Solve all problems symbolically before plugging in numbers. This will provide insight into
which variables affect your answer. It will also help avoid excessive rounding error.

Kinematics
The most abused formulas in kinematics are the definitions vaverage = Δx/Δt and v = Δx/Δt
in the limit of Δt goes to zero. The first formula can be used correctly, though vaverage doesn’t
appear anywhere in the kinematics formulas and is therefore not particularly useful. It is
useful when acceleration is zero. The second formula can only be used correctly when using
calculus.

The symbol g is used for the magnitude of the free fall acceleration due to gravity. It is
always positive since it is a magnitude, so don’t ever write, “g = -9.8 m/s2.” It varies from
place to place, but the typical textbook uses 9.81 m/s2. We use 9.806 m/s2 in labs in Salem,
Oregon. The acceleration (vector) of an object under the influence of gravity alone using a
conventional coordinate system will be (0, -g).

A common misunderstanding is that positive acceleration is speeding up and negative
acceleration is slowing down. This is not necessarily true. You need to look at the
relationship between the velocity and acceleration vectors to determine this. If the velocity
and acceleration vectors are in the same direction, then the object is speeding up. A dropped
ball is an example of negative acceleration and increasing speed. If the velocity and
acceleration vectors are in the opposite direction, then the object is slowing down. A car
driving to the left and applying the brakes has a negative velocity and a positive acceleration,
yet is slowing down. If the velocity and acceleration are consistently perpendicular, then
there is no change in speed. This occurs in uniform circular motion.

Forces
Generally, start your problem with a free body diagram for each object. It is best practice to
identify the type, direction, source, and receiver for each force. The source of the force
should be the agent that directly acts upon the object. For weight forces, this is almost
always the earth. All mathematical relationships should be consistent with your free body
diagram and a choice of Newton’s first or second law.

Newton’s first law states if ΣF = 0 then v will be constant and a = 0. It also works in
reverse: the observation of v being constant leads to the conclusion that ΣF = 0.
Newton’s second law gives a mathematical relationship between the net force on an object,
the mass of an object, and the acceleration of an object as follows: ΣF = ma.

Newton’s third law states: Fab = -Fba. The forces are of the same type with the source and
receiver reversed. Do not make the mistake of calling a normal force and a weight force
acting on the same object an action-reaction pair. Though they may be equal and opposite
forces, they are not the same type of force and they do not act on different objects.

The weight force on an object near the surface of the earth has a magnitude of m*g and a
direction of towards the center of the earth. See the notes on g in the kinematics section.

If you are not sure which type of force is exerted, it might be a normal force or tension
force. For macroscopic objects, pushing forces are typically normal forces while pulling
forces are typically tension forces.

Do not write N = mg or T = mg unless you have a good reason. A good reason is if those
are the only two forces in the y direction and there is no acceleration in that direction. You
typically use Newton’s first or second law to calculate N or T. There are no general formulas
for these forces. The direction of the normal force is always perpendicular to the surface and
repulsive. The direction of the tension force is always attractive.

The source of a tension force is the string, rope, etc. Do not attribute a tension force to the
object at the other end of the string, rope, etc.

Make a distinction between a force of friction, fk or fs, and a coefficient of friction, μk or μs.

Do not write fk = μkmg unless you already know that N = mg. The correct formula is fk =
μkN. Use Newton’s first or second law to calculate N first, and then plug it into the formula
for friction.

Do not write fs = μsN unless you have a good reason. A good reason is if the problem calls
for the maximum static friction. The correct formula is fs ≤ μsN.

For friction problems, use the following method:

1) Does the problem state definitively if the surface of the object is moving in relation to
   the other surface? If it is moving, then use kinetic friction (3). If it is not necessarily
   moving, then use static friction (2). Note that the surface of an object that is rolling
   without sliding is not moving in relation to the surface of the ground.
2) Test if the other forces can “break” the static friction by figuring out (free body
   diagrams, Newton’s first or second law, and vector addition) what magnitude the friction
   needs to be to hold the object in place relative to the surface, and then comparing that
   value to the maximum possible static friction, μsN. If the static frictional force can hold
   the object in place, then the static force of friction will take the magnitude and direction
   necessary to do so. If the static frictional force cannot hold the object in place, then the
   object will move, so you must use kinetic friction (3).
3) If the other forces break the static friction or the object is already known to be moving
   relative to the surface, simply use the formula fk = μkN for the magnitude. The direction
   will be opposite the motion of the object’s surface in relation to the other surface.

Frictional forces always acts parallel to the contact surface and perpendicular to the normal
force. The static frictional force will take the magnitude (up to a maximum) and direction
necessary to hold the object in place relative to the surface. The kinetic frictional force will
always have a magnitude of μkN and a direction opposite the relative motion.

Note that the formulas for friction are empirical. They cannot be derived from first
principles of physics. Don’t bet your life on them and don’t expect experiment to always
match theory.

Vector Components
When taking components of force vectors, the following formulas are used when the angle is
measured counterclockwise from the positive x axis:
Fx = F*cosθ
Fy = F*sinθ
The common exception to these formulas is when you use a rotated coordinate system such
as in the sliding block problem. In this case, the angle is often measured from the negative y
axis, so the trig functions get switched and you must put in a + or – sign manually.

Circular Motion
If an object moves in a circle with constant speed, then it is characterized as having uniform
circular motion. The object will be accelerating towards the center of the circle (centripetal
acceleration) with the following magnitude: acp = v2/r. The so-called centripetal force is the
net force on the object which will be macp. The centripetal force is not a new force to add to
the list of forces such as N, W, etc.; it is a force or combination of forces whose vector sum
points towards the center of a circle.

If an object or point on an object moves in a circle with changing speed, then things get
more complicated. If the angular acceleration (α) is constant, then the rotational kinematics
equations as listed in the text apply. There are specific mathematical relationships between
the angular and linear quantities. The tangential speed of a point on a rotating object is v =
rω. The centripetal acceleration of a point on a rotating object is the same as that for
uniform circular motion: acp = v2/r. This formula can be rewritten as acp = rω2. This is only
one component of the acceleration vector. The other component is the tangential
acceleration atan = rα.

Energy
Many problems that are difficult to solve or cannot be solved with Newton’s laws can often
be solved using an energy calculation. This is often when forces vary in magnitude or
direction as a function of time. The use of energy methods can come at the expense of never
knowing time.
1) Specifically identify the starting and ending events for a particular system.
2) Write down the “macroscopic energy accounting” equation: ΣUi + ΣKi + ΣW = ΣUf +
ΣKf.
3) Specifically identify the forces (type and source) affecting the objects at any time between
the starting and ending events.
4) Decide how to take into account the effect of each force on the equation. Each force
must be categorized as having a work function of zero, a potential energy function, or a non-
zero work function.
5) Calculate the work done by all non-conservative forces with W = ∫F∙ds. This can be
simplified to W = F∙s = Fs*cosφ when the force is constant and the direction of
displacement is always the same. It is helpful to be familiar with special cases where the force
and displacement vectors are parallel (φ = 0), anti-parallel (φ = 180°), or perpendicular (φ =
90°).
6) Insert formulas for U (depends on the type of force) and K (½mv2).
7) Solve for the variable of interest.

Do make a clear distinction between when the problem asks for the work done by a
particular force and the total work. The formula W = F*d*cosφ refers to the work done by a
particular force. The formula Wtotal = Fnet*d*cosφ refers to the total work. The formula W =
-ΔU refers to the relationship between the work done by a particular force and the change in
potential energy associated with that same force. It is not a useful formula unless you are
deriving a potential energy function (you won’t do this). The formula Wtotal = ΔK refers to
the total work by all forces.

Do think very carefully about the angle in the formula W = F*d*cosφ. This is the angle
measured between the force and the displacement. This is neither the angle of the force nor
the angle of the displacement. The numbers F and d are magnitudes and they are always
positive. The term cosφ will determine the sign of the work.

Do not omit or double count anything from the formula: ΣUi + ΣKi + ΣW = ΣUf + ΣKf.
Any of the five terms might be zero, or they might have more than one term. This formula is
a more refined version of the formula Wtotal = ΔK.

Momentum
The definition of momentum is p = mv. This is a useful quantity to analyze when objects
interact with one another with strong forces that cannot easily be calculated, such as in a
collision or explosion.

Total momentum for an object or system of objects will be conserved (no change) if the net
external force is zero or insignificant compared to the internal forces. The relationship can
be written mathematically as: Σpi = Σpf. Since momentum is a vector, this should be broken
up into x and y components when the problem has 2 dimensions. In one dimension, the sign
of the velocity should indicate the direction of motion.

There are certain types of collisions which are special. A collision (not an object) can be
characterized as completely inelastic, elastic, or neither. If the collision is completely inelastic,
then the velocities of all objects will be the same after the collision (the objects stick
together). If the collision is elastic, then the total kinetic energy of the system does not
change: ΣKi = ΣKf. Do not assume that a collision must be completely inelastic or elastic.
Rotational Kinematics
The most misused formula in linear kinematics is vave = Δx/Δt. Similarly, the most misused
formula in rotational kinematics is ωave = Δθ/Δt. This is a true formula, but the variable ωave
is not in any of the other rotational kinematics formulas and is rarely a useful quantity to
calculate.

Something that is never true:
g = -9.8 m/s2

Things to use only if you are absolutely sure they are true and useful:
vave = Δx/Δt (True, but seldom useful. I refer to this formula as “6th grade kinematics”.)
ωave = Δθ/Δt (True, but seldom useful. This would be “6th grade rotational kinematics.”)
N = mg (True only in a special case)
N = mg*cos(theta) (True only in a special case)
T = mg (True only in a special case)
Work = F*d (True only in a special case)
fs = μsN (True only in a special case)

								
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