


Problem Solving with Newton’s Laws
Dynamics problem solving in physics class usually involves
difficult calculations that take into account a number of vectors
on a freebody diagram. SAT II Physics won’t expect you to make
any difficult calculations, and the test will usually include the
freebody diagrams that you need. Your task will usually be to interpret
freebody diagrams rather than to draw them.
Example 1

The figure above gives us a freebody diagram
that shows us the direction in which all forces are acting, but
we should be careful to note that vectors in the diagram are not
drawn to scale: we cannot estimate the magnitude of C simply
by comparing it to M and L.
What is the magnitude of the force Curly is exerting?
Since we know that the motion of the sled is in the direction, the net force, M + L + C, must
also be in the direction. And since
the sled is not moving in the direction, the ycomponent
of the net force must be zero. Because the ycomponent
of Larry’s force is zero, this implies:
where is the ycomponent
of M and is the ycomponent of C.
We also know:
If we substitute these two equations for and into the equation , we have:
What is the acceleration of the sled?
According to Newton’s Second Law, the acceleration of
the sled is a = F/m.
We know the sled has a mass of 10 kg, so we just need
to calculate the magnitude of the net force in the direction.
Now that we have calculated the magnitude of the net force
acting on the sled, a simple calculation can give us the sled’s
acceleration:
We have been told that the sled is moving in the direction, so the acceleration is also
in the direction.
This example problem illustrates the importance of vector
components. For the SAT II, you will need to break vectors into
components on any problem that deals with vectors that are not all
parallel or perpendicular. As with this example, however, the SAT
II will always provide you with the necessary trigonometric values.
Example 2

The answer to question 1 is B. The two forces
in that diagram cancel each other out, so the net force on the particle
is zero. The velocity of a particle only changes under the influence of
a net force. The answer to question 2 is C. The net
force is in the same direction as the particle’s motion, so the
particle continues to accelerate in the same direction. The answer to
question 3 is A. Because the force is acting perpendicular
to the particle’s velocity, it does not affect the particle’s speed,
but rather acts to pull the particle in a circular orbit. Note, however,
that the speed of the particle only remains constant if the force
acting on the particle remains perpendicular to it. As the direction
of the particle changes, the direction of the force must also change
to remain perpendicular to the velocity. This rule is the essence of
circular motion, which we will examine in more detail later in this
book. The answer to question 4 is D. The net force
on the particle is in the opposite direction of the particle’s motion,
so the particle slows down, stops, and then starts accelerating
in the opposite direction.
