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
This preview has intentionally blurred sections. Sign up to view the full version.
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
)]
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 Goodstein
 Angular Momentum, Force, Mass

Click to edit the document details