9781118230725.pdf

5-1 NEWTON’S FIRST AND SECOND LAWS 97

That assumption works well if, say, a puck is sent sliding along a shortstrip
of frictionless ice — we would find that the puck’s motion obeys Newton’s laws.
However, suppose the puck is sent sliding along a longice strip extending from
the north pole (Fig. 5-2a). If we view the puck from a stationary frame in space,
the puck moves south along a simple straight line because Earth’s rotation
around the north pole merely slides the ice beneath the puck. However, if we
view the puck from a point on the ground so that we rotate with Earth, the
puck’s path is not a simple straight line. Because the eastward speed of the
ground beneath the puck is greater the farther south the puck slides, from our
ground-based view the puck appears to be deflected westward (Fig. 5-2b).
However, this apparent deflection is caused not by a force as required by
Newton’s laws but by the fact that we see the puck from a rotating frame. In this
situation, the ground is a noninertial frame,and trying to explain the deflection
in terms of a force would lead us to a fictitious force. A more common example
of inventing such a nonexistent force can occur in a car that is rapidly increas-
ing in speed. You might claim that a force to the rear shoves you hard into the
seat back.
In this book we usually assume that the ground is an inertial frame and that
measured forces and accelerations are from this frame. If measurements are made
in, say, a vehicle that is accelerating relative to the ground, then the measurements
are being made in a noninertial frame and the results can be surprising.
Figure 5-2(a) The path of a puck sliding
from the north pole as seen from a station-
ary point in space. Earth rotates to the east.
(b) The path of the puck as seen from the
ground.

N

S

W E

(a)

(b)

Earth's rotation

causes an

apparent deflection.

Checkpoint 1

Which of the figure’s six arrangements correctly show the vector addition of forces

and to yield the third vector, which is meant to represent their net force F?

:

F net

:

2

F

:

1

(a) ((b) c)

F 1 F 1 F 1

F 1 F 1 F 1

F 2 F 2 F 2

F 2

(d) ((e) F 2 f) F 2

Mass

From everyday experience you already know that applying a given force to bod-
ies (say, a baseball and a bowling ball) results in different accelerations. The com-
mon explanation is correct: The object with the larger mass is accelerated less.
But we can be more precise. The acceleration is actually inversely related to the
mass (rather than, say, the square of the mass).
Let’s justify that inverse relationship. Suppose, as previously, we push on the
standard body (defined to have a mass of exactly 1 kg) with a force of magnitude
1 N. The body accelerates with a magnitude of 1 m/s^2. Next we push on body X
with the same force and find that it accelerates at 0.25 m/s^2. Let’s make the (cor-
rect) assumption that with the same force,

mX

m 0



a 0

aX

,