sanchez (ds28677) – homework 09 – Turner – (58220)
1
This
printout
should
have
13
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A block is pushed upward along a frictionless
inclined plane.
m
F
θ
Determine the horizontal force
F
on the
block which causes it to move upward with a
constant velocity.
1.
F
=
m g
cos
θ
2.
F
=
m g
cot
θ
3.
F
=
m g
cos
θ
4.
F
=
m g
tan
θ
correct
5.
F
=
m g
sin
θ
6.
F
=
m g
sin
θ
Explanation:
The block is subject to three forces:
its
weight
vector
W
, the normal force
vector
N
of the surface,
and the external force
vector
F
which pushes it to
the right. The external force acts in the hor
izontal direction, the weight acts vertically,
and the normal force acts in the direction nor
mal (perpendicular) to the surface — leftward
from vertically up by angle
θ
. Let the
x
axis
point horizontally to the left while the
y
axix
points vertically up. In these coordinates, the
net force on the block has components
F
net
x
=
F
−
N
sin
θ,
F
net
y
=
N
cos
θ
−
W.
Since the block does not accelerate, both com
ponents must vanish, hence
N
=
W
cos
θ
=
mg
cos
θ
and
F
=
N
sin
θ
=
mg
sin
θ
cos
θ
=
mg
tan
θ.
002
(part 1 of 2) 10.0 points
The 6
.
9 N weight is in equilibrium under the
influence of the three forces acting on it. The
F
force acts from above on the left at an angle
of
α
with the horizontal. The 6 N force acts
from above on the right at an angle of 57
◦
with
the horizontal. The force 6
.
9 N acts straight
down.
F
6 N
6
.
9 N
57
◦
α
What is the magnitude of the force
F
?
Correct answer: 3
.
76405 N.
Explanation:
Standard angular measurements are from
the positive
x
axis in a counterclockwise di
rection.
Let :
F
1
=
F ,
F
2
= 6 N
,
α
2
= 57
◦
,
and
F
3
=
−
6
.
9 N
.
Consider the free body diagram. The green
vectors are the components of the slanted
forces.
F
1
F
2
F
3
α
2
α
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
sanchez (ds28677) – homework 09 – Turner – (58220)
2
The weight is is equilibrium, so
summationdisplay
F
x
=
F
1
x
+
F
2
x
+
F
3
x
= 0
F
1
x
=
−
F
2
cos
α
2
−
0
=
−
(6 N) cos 57
◦
=
−
3
.
26783 N
and
summationdisplay
F
y
=
F
1
y
+
F
2
y
+
F
3
y
= 0
F
1
y
=
−
F
2
sin
α
2
−
F
3
=
−
(6 N) sin 57
◦
−
(
−
6
.
9 N)
= 1
.
86798 N
,
and
F
1
=
radicalBig
F
2
1
x
+
F
2
1
y
=
radicalBig
(
−
3
.
26783 N)
2
+ (1
.
86798 N)
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Physics, Force, Friction, Work, kg, Sanchez

Click to edit the document details