 1 
Physics 112  HW #4 Solutions
Fall 2005
426
[Step on it!]
Assuming the truck is accelerating to the right (
→
),
as shown:
(a)
Forces on Box:
Forces on Truck:
f
1
N
1
w
=
m
g
B
B
(normal
contact force
due to Truck)
(friction
contact force
due to Truck)
(weight due
to Earth)
f
2
N
2
N
R
f
R
w
=
m
g
T
T
(normal
contact force
due to Box)
(friction
contact force
due to Box)
(friction
contact force
due to Road)
(drag force
due to Air
and Road)
f
d
(weight due
to Earth)
(normal
contact force
due to Road)
NOTE
CAREFULLY!:
*
The kinetic (sliding) friction forces
f
1
and f
2
are in the directions shown because the bottom
surface of the Box is sliding to the left (
←
)
relative to the surface of the Truck
with which it is in
contact.
Remember, the accelerating truck is
not
an inertial reference frame.
*
N
2
, the normal contact force
acting on the Truck
due to the Box, is NOT the same thing as the
weight of the Box, which is the gravitational force
acting on the Box
due to the Earth.
These are
conceptually different forces for different interactions
and must be treated as such!
(b)
There are two 3rd Law interaction pairs among the forces shown above:
(i)
N
1
and N
2
(normal contact interaction between the Box and Truck)
(ii)
f
1
and f
2
(frictional contact interaction between the Box and Truck)
a
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431
[Force is a Vector]
We need to sum vectors
F
1
and
F
2
to form the vector
R
.
F
1
+
F
2
+
R
x components:
F
2
cos
θ
= R
(1)
y components:
F
2
sin
θ
– F
1
= 0,
or
F
2
sin
θ
= F
1
(2)
Divide eq (2) by eq (1):
tan
θ
=
F
1
R
F
1
/R =
1300 N
1300 N
= 1
So
θ
= 45
°
and the angle between
F
1
and
F
2
is
135
°
Now use eq (1):
F
2
cos
θ
= R
F
2
=
R
cos
θ
=
1300 N
cos 45
°
=
1840 N
(Note: drawing a good vector diagram is the key to solving this type of problem easily.)
439
[Pulling Crates]
(a)
2.50 m/s
2
(Both crates move together.)
T
N
1
m
g
1
(normal
contact force
due to surface)
(tension
force due
to rope
(weight due
to Earth)
+x
+y
T
N
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 Fall '07
 LECLAIR,A
 mechanics, Force, Friction, contact force, normal contact force

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