101
5-1 NEWTON’S FIRST AND SECON D L AWS
x components: Along the x axis we have
F 3,x ϭ ma x Ϫ F 1, x Ϫ F 2,x
ϭ m(a cos 50Њ) Ϫ F 1 cos(Ϫ150Њ) Ϫ F 2 cos 90Њ.
Then, substituting known data, we find
F 3,x ϭ (2.0 kg)(3.0 m/s
2
) cos 50Њ Ϫ (10 N) cos(Ϫ150Њ)
Ϫ (20 N) cos 90Њ
ϭ 12.5 N.
y components: Similarly, along the y axis we find
F 3, y ϭ ma y Ϫ F 1, y Ϫ F 2, y
ϭ m(a sin 50Њ) Ϫ F 1 sin(Ϫ150Њ) Ϫ F 2 sin 90Њ
ϭ (2.0 kg)(3.0 m/s
2
) sin 50Њ Ϫ (10 N) sin(Ϫ150Њ)
Ϫ (20 N) sin 90Њ
ϭ Ϫ10.4 N.
Vector: In unit-vector notation, we can write
ϭ F 3, x ϩ F 3, y ϭ (12.5 N) Ϫ (10.4 N)
Ϸ (13 N) Ϫ (10 N) .
(Answer)
We can now use a vector-capable calculator to get the magnitude and the angle of . We can also use Eq. 3-6 to obtain
the magnitude and the angle (from the positive direction of
the x axis) as
and
(Answer)
ϭ tan
Ϫ1
F 3,y
F 3, x
ϭ Ϫ40Њ.
F 3 ϭ 2F 3,x
2 ϩ F
2
3,y ϭ 16 N
F
:
3
j
ˆ
i
ˆ
j
ˆ
i
ˆ
j
ˆ
i
ˆ
F
:
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 accelerated at 3.0 m/s
2
in the direction shown by , over a frictionless horizontal surface. The acceleration is caused by three
horizontal forces, only two of which are shown: of magnitude 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,
we cannot find
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 x axis and then along
the y axis. Caution: Use only one axis at a time.
F
:
2
ϪF
:
2
F
:
1
ϪF
:
1
ma
:
F
:
3
F 3
: ϭ ma
: Ϫ F
:
1 Ϫ F 2
: .
F
:
1 ϩ F 2
: ϩ F 3
: ϭ ma
:
(F
:
net ϭ ma
: )
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
:
3
ϪF
:
2
ϪF
:
1
ma
:
F
:
3
a
:
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.
5-1 NEWTON’S FIRST AND SECON D L AWS
x components: Along the x axis we have
F 3,x ϭ ma x Ϫ F 1, x Ϫ F 2,x
ϭ m(a cos 50Њ) Ϫ F 1 cos(Ϫ150Њ) Ϫ F 2 cos 90Њ.
Then, substituting known data, we find
F 3,x ϭ (2.0 kg)(3.0 m/s
2
) cos 50Њ Ϫ (10 N) cos(Ϫ150Њ)
Ϫ (20 N) cos 90Њ
ϭ 12.5 N.
y components: Similarly, along the y axis we find
F 3, y ϭ ma y Ϫ F 1, y Ϫ F 2, y
ϭ m(a sin 50Њ) Ϫ F 1 sin(Ϫ150Њ) Ϫ F 2 sin 90Њ
ϭ (2.0 kg)(3.0 m/s
2
) sin 50Њ Ϫ (10 N) sin(Ϫ150Њ)
Ϫ (20 N) sin 90Њ
ϭ Ϫ10.4 N.
Vector: In unit-vector notation, we can write
ϭ F 3, x ϩ F 3, y ϭ (12.5 N) Ϫ (10.4 N)
Ϸ (13 N) Ϫ (10 N) .
(Answer)
We can now use a vector-capable calculator to get the magnitude and the angle of . We can also use Eq. 3-6 to obtain
the magnitude and the angle (from the positive direction of
the x axis) as
and
(Answer)
ϭ tan
Ϫ1
F 3,y
F 3, x
ϭ Ϫ40Њ.
F 3 ϭ 2F 3,x
2 ϩ F
2
3,y ϭ 16 N
F
:
3
j
ˆ
i
ˆ
j
ˆ
i
ˆ
j
ˆ
i
ˆ
F
:
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 accelerated at 3.0 m/s
2
in the direction shown by , over a frictionless horizontal surface. The acceleration is caused by three
horizontal forces, only two of which are shown: of magnitude 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,
we cannot find
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 x axis and then along
the y axis. Caution: Use only one axis at a time.
F
:
2
ϪF
:
2
F
:
1
ϪF
:
1
ma
:
F
:
3
F 3
: ϭ ma
: Ϫ F
:
1 Ϫ F 2
: .
F
:
1 ϩ F 2
: ϩ F 3
: ϭ ma
:
(F
:
net ϭ ma
: )
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
:
3
ϪF
:
2
ϪF
:
1
ma
:
F
:
3
a
:
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.