homework 08 – RAHMAN, TARIQUE – Due: Feb 20 2008, 3:00 am
1
Question 1, chap 6, sect 1.
part 1 of 2
10 points
A block of mass
m
is accelerated across a
rough surface by a force of magnitude
F
that
is exerted at an angle
φ
with the horizontal,
as shown above. The frictional force on the
block exerted by the surface has magnitude
f
.
f
F
φ
m
What is the magnitude of the acceleration
va
of the block?
1.

va

=
F
m
2.

va

=
F
cos
φ
−
f
m
correct
3.

va

=
F
−
f
m
4.

va

=
F
cos
φ
m
5.

va

=
F
sin
φ
−
mg
m
Explanation:
From Newton’s second law of motion, the
acceleration is the total force in the horizontal
direction divided by the mass. There are two
forces in the horizontal direction: one is the
friction force; the other is the horizontal com
ponent of the dragging force
F
, but they are
in the opposite directions, so the acceleration
of the block is

va

=
F
cos
φ
−
f
m
.
Question 2, chap 6, sect 1.
part 2 of 2
10 points
Which of the following expressions for the
coe±cient of friction is correct?
1.
μ
=
mg
f
2.
μ
=
f
mg
−
F
cos
φ
3.
μ
=
f
mg
−
F
sin
φ
correct
4.
μ
=
f
mg
5.
μ
=
mg
−
F
cos
φ
f
Explanation:
By de²nition, the coe±cient of kinetic fric
tion is the ratio of the friction force and the
normal force in the vertical direction. And
it is easy to see that the normal force in the
vertical direction is just
N
=
mg
−
F
sin
φ,
So the coe±cient of friction is
μ
=
f
mg
−
F
sin
φ
.
Question 3, chap 6, sect 1.
part 1 of 3
10 points
A block is at rest on the incline shown
in the ²gure. The coe±cients of static and
kinetic friction are
μ
s
= 0
.
87 and
μ
k
= 0
.
74,
respectively.
The acceleration of gravity is 9
.
8 m
/
s
2
.
27 kg
μ
37
◦
What is the frictional force acting on the
27 kg mass?
Correct answer: 159
.
24 N (tolerance
±
1 %).
Explanation:
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documenthomework 08 – RAHMAN, TARIQUE – Due: Feb 20 2008, 3:00 am
2
F
f
N
mg
37
◦
The forces acting on the block are shown
in the ±gure. Since the block is at rest, the
magnitude of the friction force should be equal
to the component of the weight on the plane
of the incline
F
f
=
M g
sin
θ
= (27 kg) (9
.
8 m
/
s
2
) sin 37
◦
= 159
.
24 N
.
Question 4, chap 6, sect 1.
part 2 of 3
10 points
What is the largest angle which the incline
can have so that the mass does not slide down
the incline?
Correct answer: 41
.
0233
◦
(tolerance
±
1 %).
Explanation:
The largest possible value the static friction
force can have is
F
f,max
=
μ
s
N
, where the
normal force is
N
=
Mg
cos
θ
. Thus, since
F
f
=
M g
sin
θ
,
M g
sin
θ
m
=
μ
s
M g
cos
θ
m
tan
θ
m
=
μ
s
θ
m
= tan
−
1
(
μ
s
)
= tan
−
1
(0
.
87)
= 41
.
0233
◦
.
This is the end of the preview. Sign up
to
access the rest of the document.