This preview shows pages 1–3. Sign up to view the full content.
48
Chapter 4
EVALUATE:
The friction force on the box, exerted by the bed of the truck, is in the direction of the truck’s
acceleration. This friction force can’t be large enough to give the box the same acceleration that the truck has and
the truck acquires a greater speed than the box.
Figure 4.30
4.31.
IDENTIFY:
Identify the forces on the chair. The floor exerts a normal force and a friction force.
SET UP:
Let
be upward and let
y
+
x
+
be in the direction of the motion of the chair.
EXECUTE:
(a)
The freebody diagram for the chair is given in Figure 4.31.
(b)
For the chair,
so
0
y
a
=
y
y
Fm
a
=
∑
gives
sin37
0
nm
gF
−
−°
=
and
142 N
n
=
.
EVALUATE:
n
is larger than the weight because
F
!
has a downward component.
Figure 4.31
4.32.
IDENTIFY:
Identify the forces on the skier and apply
m
=
∑
F
a
!
!
. Constant speed means
.
0
a
=
SET UP:
Use coordinates that are parallel and perpendicular to the slope.
EXECUTE:
(a)
The freebody diagram for the skier is given in Figure 4.32.
(b)
x
x
a
=
∑
with
gives
.
0
x
a
=
2
sin
(65.0 kg)(9.80 m/s )sin 26.0
279 N
Tm
g
θ
==
°
=
EVALUATE:
T
is less than the weight of the skier. It is equal to the component of the weight that is parallel to the
incline.
Figure 4.32
4.33.
IDENTIFY:
m
∑
F
=a
!
!
must be satisfied for each object. Newton&s third law says that the force
that the car
exerts on the truck is equal in magnitude and opposite in direction to the force
C on T
F
!
T on C
F
!
that the truck exerts on the
car.
SET UP:
The only horizontal force on the car is the force
T on C
F
!
exerted by the truck. The car exerts a force
on the truck. There is also a horizontal friction force
C on T
F
!
f
!
that the highway surface exerts on the truck. Assume
the system is accelerating to the right in the freebody diagrams.
EXECUTE:
(a)
The freebody diagram for the car is sketched in Figure 4.33a
(b)
The freebody diagram for the truck is sketched in Figure 4.33b.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentNewton&s Laws of Motion
49
(c)
The friction force
f
!
accelerates the system forward. The tires of the truck push backwards on the highway
surface as they rotate, so by Newton&s third law the roadway pushes forward on the tires.
EVALUATE:
and
each equal the tension
T
in the rope. Both objects have the same acceleration
T on C
F
C on T
F
a
!
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Acceleration, Force, Friction

Click to edit the document details