c. 1
E = stress/strain
Detennine the
elastic modulus.
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a.
1
Apply the bulk modulus
formula and detennine the
change in volume of the
water. Note: assume that
all of the air in the
container was removed
before it was lowered.
Solution: (Section 9-6)
D.
V = - 1.5
X
10-6 m3
or
- 1.5 cm3
The change in volume is negative because the water is compressed
as the pressure increases.
b.l
% change = (- 1.5 cm3)/(1000
Determine the %
change in the volume.
% change = - 0.15%
c. 1
To a good approximation the phrase is correct.
Is the phrase "water is
incompressible" correct?
9-17
cm3)
x 100%
a.3
Apply the second
condition of equilibrium,
~ l' = 0, and solve for
the tension in the
wire. Note: choose the
point where the pole
is attached to the wall as
the rotation point.
The convenient point for the axis of rotation would be a point where
an unknown force is located. Therefore, -either end of the pole would
be a convenient point. Let us choose the end of the pole that meets
the wall as the rotation point. At this point the wall exerts a force
which is represented by the components, H and V.
Use the point of pencil method to determine the direction of each
torque. From the diagram it can be seen that T produces a CCW
torque while the 800 N weight of the pole and the 400 N load each
produce a CW torque. Since V and H act at the rotation point, the
lever arm distance is zero and they produce no torque.
1------------------------------------,
I
400 N
,
,
,
800 N
rotation
,
,
------------------------------------------2.4 m
point -
cw,'
f-
4.8 m
cw",/
,
..j
T(6.00m)(sin 37°)(sin 90°) - (800 N)(3.00 m)(cos 37°)(sin 90°)
- (400 N)(6.00 m)(cos 37°)(sin 90°) = 0
(3.60 m) T - 1920 m N - 1920 m N = 0
T = 1070 N
b. 1
The value of V was
obtained in step a. 2.
T is now known; solve
for H.
From step a. 2
H-T=O
H - 1070 N = 0
H = 1070 N
9-15
a.
1
Draw an accurate
free body diagram
locating the
forces acting on
the plank.
a.
2
Apply the first condition
of equilibrium, 1: F = O.
Solution: (Section 9-3)
Let F represent the upward force that the edge of the roof exerts on
the plank and the man.
1:Fx = 0
No horizontal forces are present in this problem.
1:Fy = 0
F - (150 kg)(9.8 rnjs2) - (70 kg)(9.8 rnjs2) = 0
F = 2160 N
a.3
Choose the left end
of the plank as the
rotation point. Use the
tip of pencil method
to determine the
direction (CWor CCW)
the torque produced
by each force.
According to theory, since no rotation occurs, any point may be
selected to be the location of the reference point. Therefore, let the
left end of the plank be chosen as the rotation point. Using the point
of pencil method, place the point of the pencil at the left end of the
line representing the plank. Pushing on the pencil at the point where
F is located and in the direction of F causes the pencil to rotate CCW.
Thus F causes a counterclockwise torque about the about the left end
of the plank. Now repeat for the stunt man's weight and the weight of
of the board. Arrange the pencil so that the stunt man's weight and the
weight of the board act. Note that each force produces a clockwise
torque (CW) about the left end of the board.
a.4
Use the second
condition of
equilibrium to solve
for length of
overhang (x).
(2160 N)(10.0 m - x) sin 90° - (1470 N)(5.00 m) sin 90°
- (686 N)(10.0 m) sin 90° = 0
21,600 Nm - (2160 N) x - 7350 Nm - 6860 Nm = 0
7390 Nm - (2160 N) x = 0
x = (7390 Nm)/2160 N) = 3.42 m
a.5
Choose the edge of the
roof as the reference point.
Use the tip of pencil
method to determine the
direction of the torque
produced by each force.
Using the point of pencil method, place the point of the pencil at the
point where the plank touches the edge of the roof, Le., at point F in
the diagram. Pushing on the pencil at the point where F is located does
not produce rotation about this point. Therefore, force F does not
produce a torque about the rotation point because the lever arm
distance from F to the reference point is zero. The weight of the plank
causes a CCW rotation while the weight of the man causes a CW
rotation about the reference point.
9-13
a.2
~ Fx = 0
Apply the fIrst condition
of equilibrium, ~F = O.
~Fy = 0
a.3
Apply the second
condition of
equilibrium, d'! = O.
Choose the center of
the table as the axis
of rotation.
No horizontal forces are present in this problem.
FA + FB
-
700 N - 150 N = 0
_____~~~!J3 = ~50~_~<1uati~~J2..
_
In order to determine whether the torque produced by a force about
the rotation point is clockwise (CW) or counterclockwise (CCW),
let us use the tip of pencil method. Place the point of a pencil at
the axis of rotation and hold it fIxed. The rest of the pencil is
parallel to the table on which the forces act. Applying the force
to the pencil at the point in the diagram where the force is located
will cause the pencil to rotate either CW or CCW.
For example, place the point of the pencil at the center of the line
representing the table. Pushing on the pencil at the point where
Andy's force is located and in the direction of Andy's force
causes the pencil to rotate CWo Thus, Andy's force causes a clockwise torque about the center of the board.
Now arrange the
pencil so that Bob's force acts. Note that this force produces a
counterclockwise torque. Repeating this process we fmd that Chuck's
weight produces a CCW torque. The weight of the board creates
no torque. The weight is at the rotation point. The lever arm
distance is zero; therefore, the torque is zero .
. -----
\
rotation point at
center of board
(
\
,\CCW
iFB
'JOO N
r- ~ccw
Since no rotation occurs,
FB
(1.0 m) + (700 N)(OAO m) - FA (1.0 m) = 0
Simplifying:
FB -
FA = - 280 N (equation 2)
9-11