A bowling ball collides with a ping-pong ball. In that collision,
which one experiences the greater force magnitude?
1.
The bowling ball.
2.
The ping-pong ball.
3.
It depends on their two speeds.
4.
None of the above.
10/26/15
Oregon State University PH 211, Class #13
1
With his Third Law of motion, Isaac Newton observed the
“double-ended” nature of any force. But he had also begun to
examine how a force—or, more generally, a vector combination
of forces, Fnet—affects the motion of any mass. He first described that in terms of its impulse—its effect upon momentum
(Fnett = P), but then he restated it more usefully.
First, notice that for any given (and constant) mass, its change in
momentum is:
P = Pf – Pi
= mvf – mvi
= m(vf – vi )
= mv
So we can re-write the basic impulse-momentum relationship:
Fnett = mv
10/26/15
Oregon State University PH 211, Class #13
2
Now just re-arrange this algebraically: Fnet = mv/t
But v/t is simply the average acceleration, aavg, of the
object over the time interval t. And if we examine
smaller and smaller increments of time, we arrive at a
powerful instantaneous predictor of an object’s motion….
Newton’s Second Law:
Fnet = dP/dt
Fnet = d(mv)/dt
For constant m:
Fnet = md(v)/dt
Fnet = ma
At any given moment in time, the acceleration, a, of any
mass, m, can be computed from the net force on that mass.
10/26/15
Oregon State University PH 211, Class #13
3
A 90-kg skater gliding over level ice is being pushed by wind at his
back with a force magnitude of 47.0 N. His skates experience a
friction force magnitude of 20 N. What is the magnitude of his
acceleration?
10/26/15
1.
0.222 m/s2
2.
0.300 m/s2
3.
0.522 m/s2
4.
3.33 m/s2
5.
None of the above.
Oregon State University PH 211, Class #13
4
Draw and completely label the correct free-body diagram (FBD) for
the previous situation (the skater blown by the wind).
This is the most useful form of visual representation for solving force
& motion problems. This is the best V in the ODAVEST procedure
when you’re analyzing forces on an object.
So… how do we “read” this diagram to get to the Equations step?
10/26/15
Oregon State University PH 211, Class #13
5
Every force is a vector, so Newton’s Second Law applies to
each vector direction individually:
Fx.net = max
Fy.net = may
or:
SFx = max
SFy = may
The acceleration values revealed here—for as long as they
remain steady—are ready-made for use in kinematic
calculations. Now we can predict any object’s motion simply
by knowing its mass and the net force acting upon it.
Apply the above to the skater FBD from the previous slide:
Write a set of correct equations (one for the x-direction, one for
the y-direction) that will let you solve for the x- and ycomponents of the skater’s acceleration.
10/26/15
Oregon State University PH 211, Class #13
6
Notice what Newton’s Second Law reveals about the force of gravity.
Consider first a projectile of mass m (and let upward be defined as
the positive y-direction):
Fx.net = max = m(0) = 0
Fy.net = may = m(–g) = –mg
The gravitational force, FG, acting on a projectile of mass m has a
magnitude of mg (and is directed downward).
Question: What is the gravitational force on an object of mass m
when it’s not a projectile?
10/26/15
Oregon State University PH 211, Class #13
7
According to Newton’s Second Law, if the net force in any direction
on a body is zero, the body will have zero acceleration in that
direction.
This specific case is important enough to note with its own law….
Newton’s First Law:
“An object’s motion will continue unchanged
unless it is acted upon by a non-zero net force.”
10/26/15
Oregon State University PH 211, Class #13
8