9781118230725.pdf

5-3APPLYING NEWTON’S LAWS 111

course) and has magnitude mg(5.00 kg)(9.80 m/s^2 )49.0 N. plane and thus cannot affect the motion along the plane. (It
has no component along the plane to accelerate the box.)
We are now ready to write Newton’s second law for mo-
tion along the tilted xaxis:

The component axis the only component of the acceleration

(the box is not leaping up from the plane, which would be

strange, or descending into the plane, which would be even

stranger). So, let’s simply write afor the acceleration along the

plane. Because is in the positive xdirection and the compo-

nentmgsinuis in the negative xdirection, we next write

Tmgsinuma. (5-24)

Substituting data and solving for a, we find

a0.100 m/s^2. (Answer)

The result is positive, indicating that the box accelerates up the

inclined plane, in the positive direction of the tilted xaxis. If

T

:

Fnet,xmax.

That direction means that only a component of the force is
along the plane, and only that component (not the full force)
affects the box’s acceleration along the plane. Thus, before we
can write Newton’s second law for motion along the xaxis, we
need to find an expression for that important component.
Figures 5-15ctohindicate the steps that lead to the ex-
pression. We start with the given angle of the plane and
work our way to a triangle of the force components (they
are the legs of the triangle and the full force is the hy-
potenuse). Figure 5-15cshows that the angle between the
ramp and Fgis 90u. (Do you see a right triangle there?)
:

Next, Figs. 5-15dtofshow and its components: One com-
ponent is parallel to the plane (that is the one we want) and
the other is perpendicular to the plane.
Because the perpendicular component isperpendicular,
the angle between it and must be u(Fig. 5-15d). The com-
ponent we want is the far leg of the component right trian-
gle. The magnitude of the hypotenuse is mg(the magnitude
of the gravitational force). Thus, the component we want has
magnitudemgsinu(Fig. 5-15g).
We have one more force to consider, the normal force
FNshown in Fig. 5-15b. However, it is perpendicular to the
:

Fg

:

Fg

:

Sample Problem 5.05 Reading a force graph

Here is an example of where you must dig information out
of a graph, not just read off a number. In Fig. 5-16a, two
forces are applied to a 4.00 kg block on a frictionless floor,
but only force is indicated. That force has a fixed magni-
tude but can be applied at an adjustable angle to the posi-
tive direction of the xaxis. Force is horizontal and fixed in
both magnitude and angle. Figure 5-16bgives the horizontal
accelerationaxof the block for any given value of ufrom 0
to 90. What is the value of axforu 180 ?

KEY IDEAS

(1) The horizontal acceleration axdepends on the net hori-
zontal force Fnet,x, as given by Newton’s second law. (2) The
net horizontal force is the sum of the horizontal compo-
nents of forces and.

Calculations:The xcomponent of is F 2 because the vector
is horizontal. The xcomponent of is F 1 cos. Using these
expressions and a mass mof 4.00 kg, we can write Newton’s
second law ( m ) for motion along the xaxis as

F 1 cosuF 2 4.00ax. (5-25)

From this equation we see that when angle u 90 ,F 1 cosu
is zero and F 2 4.00ax. From the graph we see that the

F :a

:

net

F 

:

1

F

:

2

F

:

F 2

:

1

F

:

2



F

:

1

Figure 5-16(a) One of the two forces applied to a block is shown.

Its angle ucan be varied. (b) The block’s acceleration component

axversusu.

WhenF 1 is horizontal,

the acceleration is

3.0 m/s^2.

F 1

x

θ

(a)

(b)

3

2

1

(^0) 0° 90°
θ
ax
(m/s
2 )
WhenF 1 is vertical,
the acceleration is
0.50 m/s^2.
corresponding acceleration is 0.50 m/s^2. Thus,F 2 2.00 N
and Fmust be in the positive direction of the xaxis.
:
2
we decreased the magnitude of enough to make a0, the
box would move up the plane at constant speed. And if we de-
crease the magnitude of even more, the acceleration would
be negative in spite of the cord’s pull.

T

:

T

:

Additional examples, video, and practice available at WileyPLUS

From Eq. 5-25, we find that when u 0 ,

F 1 cos 02.004.00ax. (5-26)

From the graph we see that the corresponding acceleration

is 3.0 m/s^2. From Eq. 5-26, we then find that F 1 10 N.

SubstitutingF 1 10 N,F 2 2.00 N, and u 180 into

Eq. 5-25 leads to

ax2.00 m/s^2. (Answer)