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Ph1a: Solution to the Final Exam
Alejandro Jenkins, Fall 2004
Problem 1 (10 points)  The Delivery
A crate of mass
M
, which contains an expensive piece of scientiﬁc equipment, is being
delivered to Caltech. The delivery truck has a freight bed of length
L
(see the ﬁgure), with
a coeﬃcient of static friction
μ
s
and a coeﬃcient of kinetic friction
μ
k
.
Rather than move the heavy crate himself, the driver tilts the truck bed by an angle
θ
(see ﬁgure) and then drives the truck forward with increasing acceleration
a
, until the crate
begins to slide.
M
θ
x
y
a
L
g
µ
, µ
s
k
For this problem, use the
x, y
coordinate system shown in the ﬁgure. These coordinates
are ﬁxed with respect to the truck’s bed,
not
to the ground.
(a) (2 points)
Draw a freebody force diagram for the crate in the truck’s frame of
reference.
Solution:
See Figure 1.
N
is the normal force that the truck’s bed exerts on
the crate, and
f
is the corresponding friction.
θ
x
y
f
Mg
N
Ma
Figure 1: Freebody diagram for the crate in the truck’s frame of reference
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View Full Document(b) (3 points)
Write down Newton’s second law for the motion of the crate in the
x

and
y

directions, just before it begins to slip.
Solution:
While the crate remains on the truck’s bed,
a
y
= 0. Thus:
Ma
y
=
N

Mg
cos
θ
+
Ma
sin
θ
= 0
.
The maximum static friction is given by
f
max
=
μ
s
N
. This must be the magni
tude of the friction just before the crate begins to slip. Therefore:
Ma
x
=
f
max

Mg
sin
θ

Ma
cos
θ
=
μ
s
N

Mg
sin
θ

Ma
cos
θ
= 0
.
(c) (2 points)
Determine the minimum acceleration
a
min
for which the crate will begin
to slip. Express your answer in terms of the constants shown in the ﬁgure.
Solution:
If we solve for
a
in the equations obtained in part
(b)
for the crate
just before it begins to slip, the result must correspond to
a
min
. Therefore
a
min
=
g
μ
s
cos
θ

sin
θ
μ
s
sin
θ
+ cos
θ
.
When the truck reaches
a
min
and the driver notices the crate beginning to slide, he
continues at that constant acceleration.
(d) (3 points)
Find the speed of the crate along the
x

direction when the crate leaves
the truck bed. Neglect the size of the crate. You may leave your answer in terms of
a
min
.
Solution:
If the crate is sliding, then the friction is kinetic rather than static.
By Newton’s second law, the acceleration is:
a
x
=
μ
k
(
g
cos
θ

a
min
sin
θ
)

g
sin
θ

a
min
cos
θ .
Since this acceleration is constant, we can ﬁnd the ﬁnal velocity of the crate by
using the equation
v
2
f
= 2

a
x

L .
Notice that, in our coordinate system,
a
x
for the sliding crate is negative, so
that

a
x

=

a
x
. Thus we have that
v
f
=
q
2
L
[
g
sin
θ
+
a
min
cos
θ

μ
k
(
g
cos
θ

a
min
sin
θ
)]
.
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 Fall '07
 Goodstein
 Mass

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