9781118230725.pdf

5-1 NEWTON’S FIRST AND SECOND LAWS 101

x components:Along the xaxis we have

F3,xmaxF1,xF2,x

m(acos 50)F 1 cos( 150 )F 2 cos 90.

Then, substituting known data, we find

F3,x(2.0 kg)(3.0 m/s^2 ) cos 50(10 N) cos( 150 )

(20 N) cos 90

12.5 N.

y components:Similarly, along the yaxis we find

F3,ymayF1,yF2,y

m(asin 50)F 1 sin( 150 )F 2 sin 90

(2.0 kg)(3.0 m/s^2 ) sin 50(10 N) sin( 150 )

(20 N) sin 90

10.4 N.

Vector:In unit-vector notation, we can write

F3,x F3,y (12.5 N) (10.4 N)

(13 N)(10 N). (Answer)

We can now use a vector-capable calculator to get the mag-

nitude and the angle of. We can also use Eq. 3-6 to obtain

the magnitude and the angle (from the positive direction of

thexaxis) as

and tan^1 (Answer)

F3,y

F3,x

 40 .

F 3  2 F3,^2 xF^2 3,y16 N

F

:

3

iˆ jˆ

F iˆ jˆ iˆ jˆ

:

3

Sample Problem 5.02 Two-dimensional forces, cookie tin

Here we find a missing force by using the acceleration. In
the overhead view of Fig. 5-4a, a 2.0 kg cookie tin is acceler-
ated at 3.0 m/s^2 in the direction shown by , over a friction-
less horizontal surface. The acceleration is caused by three
horizontal forces, only two of which are shown: of magni-
tude 10 N and of magnitude 20 N. What is the third force
in unit-vector notation and in magnitude-angle notation?

KEY IDEA

The net force on the tin is the sum of the three forces
and is related to the acceleration via Newton’s second law

. Thus,

, (5-6)

which gives us

(5-7)

Calculations: Because this is a two-dimensional problem,
wecannotfind merely by substituting the magnitudes for
the vector quantities on the right side of Eq. 5-7. Instead, we
must vectorially add , (the reverse of ), and
(the reverse of ), as shown in Fig. 5-4b. This addition can
be done directly on a vector-capable calculator because we
know both magnitude and angle for all three vectors.
However, here we shall evaluate the right side of Eq. 5-7 in
terms of components, first along the xaxis and then along
theyaxis.Caution:Use only one axis at a time.

F

:

2

F

:

F 2

:

F 1

:

ma 1

:

F

:

3

F 3

:

ma:F

:

1 F 2

:

.

F

:

1 F 2

:

F 3

:

ma:

(F

:

netma

:)

:a

F

:

net

F

:

3

F

:

2

F

:

1

:a

Additional examples, video, and practice available at WileyPLUS

Figure 5-4(a) An overhead view of two of three horizontal forces that act on a cookie

tin, resulting in acceleration. is not shown. (b) An arrangement of vectors , ,

and to find force .F

:

F 3

:

2

F

:

ma 1

F: :

a 3

:

y

(a)

30°

x

y

(b)

x

F 2

F 3

F 2

F 1

a

a

50°

m

• 
  


• F 1

These are two

of the three

horizontal force

vectors.

This is the resulting

horizontal acceleration

vector.

We draw the product

of mass and acceleration

as a vector.

Then we can add the three

vectors to find the missing

third force vector.