PHY 111 Lab 5
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Lab 5: Static Equilibrium
1.
Testable Question:
How are force in the X direction (F
X
) and force in the Y direction (F
Y
) related to the angle
(
)?
ϴ
2.
Hypothesis:
As the angle increases, F
X
will decrease and F
Y
will increase because static equilibrium
must be maintained.
3.
Variables:
Control(s):
Gravity (g), Force of the Mass (F
200
)
Independent:
Angle (
)
ϴ
Dependent:
Force in the x Direction (Fx) & Force in the Y direction (Fy)
4.
Experimental Design:
Control(s):
g = 9.81 m/s
2
, F
200
= 1.96 N
i
°
ϴ
Sin
ϴ
Cos
ϴ
M
x
(kg)
M
y
(kg)
F
x
(N)
F
y
(N)
1-10
5.
Materials:
●
Level
●
3 pulleys with clamps
●
200g mass
●
3 strings
●
Mass holder
●
Mass & hanger set
●
Force table
6. Procedure:
1.
Before the trials, the force table is set up with the 3 strings and pulleys. On one of the
strings sits the 200 g mass, and the other two strings hold 5 gram weight hangers.
2.
The X string is set at 180
°
and the Y string is set at 270
°
. The third string starts at 0
°
.
3.
Weight will be added to the X and Y strings with hangers until static equilibrium is
reached.
4.
After static equilibrium is reached, record the weight in the X and Y direction.
5.
Next, move the string with the 200g mass to 10
°
. Add or remove weight from the
hangers in the X and Y direction until static equilibrium is reached. Record the weight
added to each string.
6.
Repeat step 5, for each 10
°
increment until 90
°
is reached.
7.
Using the data, calculate the Sin
, Cos
, F
ϴ
ϴ
x
(N), and F
y
(N) for each increment.
8.
Using Microsoft Excel, plug in F
x
(N), and F
y
(N) v.
° to find the first graph.
ϴ
9.
Then, use Force in the X Direction Fx (N) v. Cos
ϴ
and Force in the Y Direction F
Y
(N)
v.
Sin
ϴ
, to create the next two graphs.
10.
The equation of the line and the R
2
value of F
x
and F
y
was found based on the scatter plot
diagram for Graph 2 and 3.
11.
The equation being investigated is
F
200
=
F
x
cosΘ
+
F
y
sinΘ
7. Data:
Control(s):
g = 9.81 m/s
2
, F
200
= 1.96 N
i
°
ϴ
Sin
ϴ
Cos
ϴ
M
x
(kg)
M
y
(kg)
F
x
(N)
F
y
(N)
1
0
0.0
1.0
0.2
0.0
1.96
0.0
2
10
0.174
0.985
0.195
0.042
1.91
0.412
3
20
0.342
0.94
0.185
0.065
1.81
0.638
4
30
0.5
0.866
0.107
0.147
1.74
1.05
5
40
0.643
0.766
0.15
0.125
1.47
1.23
6
50
0.766
0.643
0.125
0.15
1.23
1.47
7
60
0.866
0.5
0.105
0.175
1.03
1.72
8
70
0.94
0.342
0.075
0.19
0.736
1.86
9
80
0.985
0.174
0.035
0.195
0.343
1.91
10
90
1.0
0.0
0.0
0.205
0.0
2.01
8.
Analysis:
Graph-1: Force in the Y & X Direction v.
ϴ
0
10
20
30
40
50
60
70
80
90
100
0
0.5
1
1.5
2
2.5
Force in the X Direction Fx & Force in the Y Direction Fy (N) vs.
(°)
ϴ
Fx
Fy
(°)
ϴ
Force in the X Direction Fx & Force in the Y Direction
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Related Questions
1–2. Figure C1.2 shows a mechanism that is typical in the
tank of a water closet. Note that flapper C is hollow
and filled with trapped air. Carefully examine the
configuration of the components in the mechanism.
Then answer the following leading questions to gain
insight into the operation of the mechanism.
5. What effect will cause item D to move?
6. As item D is moved in a counterclockwise direction,
what happens to item F?
7. What does item F control?
8. What is the overall operation of these mechanisms?
9. Why is there a need for this mechanism and a need
to store water in this tank
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PHYS X PHYS X
印 PHYS X
PHYS X
POTPHYS X
PHYS X
E PHYS X E PHYS
top/semester2/physics%20for%20engineers/PHYS220_CH15_Lecture%20Notes_Problems%2015 19,15.29 S
(D Page view
A Read aloud
V Draw
Problem-15-19: page-475
A 0.500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion
with an amplitude of 10.0 cm. Calculate the maximum value of its
(a) speed, and acceleration.
(b) the speed and the acceleration when the object is 6.00 em from the equilibrium position, and
(c) the time interval required for the object to move from.r50 to r5 8.O0 cm.
Solution:
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PROBLEM SET # 7: THE EQUILIBRIUM MODEL
On the space provided, present correct and organized solutions to the following answered
problems. Box the final answers. Detach each page neatly and submit to your instructor.
1. A Christmas light hangs on a series of strings as shown in the figure below. The tension
on string a, Ta, is 0.48 N.
%3D
42°
b.
J
a
C
30°
25°
비 m
(a) What is the tension on string b? (Ans: 0.22 N)
(b) What is the tension on string c? (Ans: 0.35 N)
(c) What is the tension on string d? (Ans: 0.37 N)
(d) What is the mass of the Christmas light, m, in grams? (Ans: 34 grams)
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Select one:
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O b. Point functions are not the properties of the system.
O c. The Point Function depends on the path followed during a process as well as the end states.
O d. Examples for points functions are temperature, pressure, density.
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is a mass hanging by a spring under the influence of gravity. The force due to gravity, Fg, is acting
in the negative-y direction. The dynamic variable is y. On the left, the system is shown without spring deflection.
On the right, at the beginning of an experiment, the mass is pushed upward (positive-y direction) by an amount y₁.
The gravitational constant g, is 9.81 m/s².
DO
C.D
Frontly
у
Your tasks:
No Deflection
m
k
Fg = mg
Initial Condition
y
m
k
Write down an expression for the total energy If as the sum
Write down an expression for the total energy H
Fg = mg
Figure 3: System schematic for Problem 4.
Yi
&
X
Write down, in terms of the variables given, the total potential energy stored in the system when it is held in
the initial condition, relative to the system with no deflection.
as the sum of potential and kinetic energy in terms of y, y, yi
C After the system is released, it will start to move. Write down an expression for the kinetic energy of the
system, T, in terms of…
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Select one:
a. If Fe1 is negative then the spring is under compression
b. If Fe2 is negative then the spring is under tension
c. If Fe1 is positive then the spring is under tension
d. If Fe2 is positive then the spring is under tension
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1.When Dealing with liquids and gases, we ordinarily speak of stresses, for solids we speak pressures
True
False
2 .Open system is defined when a particular quantity of matter is under study. This system always contains the same matter. There can be no transfer of mass across it`s boundary
True
False
3. Thermodynamic properties can be placed in two general classes control volume and intensive
True
Flase
4. Triple point of water- is the condition of water temperature and pressure under which gaseous liquid and solid phases cannot exist in equilibrium
True
False
5. Intensive properties are additive in the sense previously considered. Their values are dependent of the size of extent of system and may vary from place to place within the system at any moment. Specific volume pressure. and temperature are important intensive properties
True
False
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In a linear spring finite element model, as per the finite element model’s sign convention, which one of the following statements is true?
Select one:
a. If Fe1 is positive then the spring is under tension
b. If Fe2 is negative then the spring is under tension
c. If Fe2 is positive then the spring is under compression
d. If Fe1 is positive then the spring is under compression
Clear my choice
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ue 9:E0
• ZAIN IQ lI.
۱۳ شباط
تحرير
QI: Find the result of F4 Force acting on a body as shown below:
400 N
300 N
60
30
F;= 150 N F,- 260 N
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• Label all forces and directions.
How many unknowns are there? (Remember, an unknown does not have to be a force. It
can also be a dimension, angle, etc.)
• How many equilibrium equations can you create?
• Is there enough information to solve for the unknowns? Explain.
●
Draw pulley A as your particle.
B
20°
A
100 N
F
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A model tow-tank test is conducted on a bare hull model at the model design
speed in calm water. Determine the effective horsepower (hp) for the ship,
including appendage and air resistances. The following parameters apply to the
ship and model:
Ship
1,100
Model
Length (ft)
Hull Wetted Surface Area (ft2)
Speed (knots)
30
250,000
15
Freshwater
Water
Seawater 50°F
70°F
Projected Transverse Area (ft²)
Cair
7,500
0.875
Appendage Resistance (% of bare hull)
10%
Hull Resistance (Ibf)
20
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is a mass hanging by a spring under the influence of gravity. The force due to gravity, Fg, is acting
in the negative-y direction. The dynamic variable is y. On the left, the system is shown without spring deflection.
On the right, at the beginning of an experiment, the mass is pushed upward (positive-y direction) by an amount y₁.
The gravitational constant g, is 9.81 m/s².
No Deflection
m
k
Fg = mg
Initial Condition
m
k
Fg = mg
Figure 3: System schematic for Problem 4.
Yi
8
Your tasks:
A Write down, in terms of the variables given, the total potential energy stored in the system when it is held in
the initial condition, relative to the system with no deflection.
B Write down an expression for the total energy H as the sum of potential and kinetic energy in terms of y, y, yi
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C After the system is released, it will start to move. Write down an expression for the kinetic energy of the
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T'sec
In helical spring experiment a
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Used the slope to find the spring
:constant, the value of k is
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
O.
O a. 13.3 N/m
O b. 6.8N/m
O c. 5 N/m
O d. 3.3 N/m
O e. 24 N/m
5
10 15 20 25 30 35 40 45 50 55 60 65
M (9)
70
75
80
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2- We can say that a beam is in equilibrium, if all forces and moments applied to it are
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Student with ODD last digits of
matric number (AD20xxxZ)
Z = 1,3,5,7,9
Student with EVEN last digits of
matric number (AD20xxxZ)
Z = 0,2,4,6,8
Parameter
X
1000 N/m
1200 N/m
Y
0.9 m
1.2 m
Z
0.6 m
0.8 m
(i)
At the point of contra-flexure, the value of bending moment is
a) Zero
b) Maximum
c) Can't be determined
d) Minimum
(ii)
its sign.
positive/negative bending moments occur where shear force changes
a) Minimum
b) Zero
c) Maximum
d) Remains same
(iii) The shear force in a beam subjected to pure positive bending
a) Positive
b) Negative
c) Zero
d) Cannot determine
(iv) SI units of shear force is
a) kN/m
b) kN-m
c) kN
d) m/N
(v)
a) Hinged
b) Simple
c) Fixed
d) Joint
support develops support moment.
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% difference between the directions of the sum of the forces due to the string 1 and 2 and the force due to string 3 = %
String Number
Mass Attached (kg)
Force Exerted (N)
Direction (°)
1
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2.45
30°
2
.25
2.45
120°
3
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3.43
255°
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=
Knowledge Check: String in a Circle
Please answer the following question(s):
1. Object of mass m tied at the end of a string of length R whirled around a circle. The tension in the
string is T. How much work does T do on the object?
V
%
Work done by T on m is
Centripetal force
Search
5
16 4-
^
6
&
♫+
7
R
because, T acts along the radius and is
to the velocity (instantaneous displacement) which is along the tangent.
do work on the object during circular motion.
0
A
8
fg
।বব
O
f10
DII
fm
DDI
f12
{1}
T
back
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themodynamic system we cannot add energy to the system, via substance (E ) (1.e. matter
which contains energy is not allowed across the boundary)
Across the Boundaries
E° = No
Q =
= Yes
W
mass NO
CLOSED
= Yes
SY STEM
m = constant
| energy YES
Figure 1.1.
If the substance inside the thermodynamic system shown in figure 1.1. (i.e. piston
cylinder device) is air, is the system a
Fixed closed system
Moveable closed system
A.
В.
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10:42
*il 64
new.edmodo.com/view-office-online,
4
2/2
7. Find the magnitude and direction of F3 in Fig. 2.7
for the given system to be in equilibrium.
F3
F = 15N
250
750
F = 30N
Fig. 2.7
+
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Select the correct response(s):
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Fluid dynamics and statics examine fluids separately.
Ooblek is an example of Newtonian Fluids.
Hydrodynamics is part of fluid statics.
Everywhere around us, there is fluid dynamics.
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The temperature of an ideal gas changes during a throttling.
True
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A Moving to another question will save this response.
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Transition
1
2
TEL L 3 14
7
A diagram of the forces being applied to a box is provided (below). If the net force acting
on the box is 10 N to the right, what is the magnitude of the force applied by the boy
pulling to the left? Record your answer in the grid below. Show all work!
Left
Right
11 Newtons
8 Newtons
2.
? Newtons
21 Newtons
Work/Math
(2)
(2
(2)
(2)
40
40
40
Answer with units
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F8
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F12
F10
000 F4
F9
F5
F6
F7
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1. For your science fair project, you decided to design a model rocket ship.
The fuel burns exerting a time-varying force on the small 2.00 kg rocket
model during its vertical launch. This force obeys the equation F= A + Bt2.
Measurements show that at t=0, the force is 25.0 N, and at the end of the
first 2.00 s, it is 45.0 N. Assume that air resistance is negligible.
a. What are the forces acting on the rocket?
b. Draw its free-body diagram.
c. Find the constants A and B, including their SI units using this
equation F= A + Bt².
d. Find the net force on this rocket and its acceleration the instant after
the fuel ignites.
e. Find the net force on this rocket and its acceleration 3.00 s after fuel
ignition.
f. Suppose you were using this rocket in outer space, far from all gravity.
What would its acceleration be 3.00 s after fuel ignition?
g. What is the rocket's mass in outer space? What is its weight?
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account_circle
Science
PhysicsQ&A LibraryA child’s toy consists of a m = 31 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monkey is first hung on the spring and the system reaches equilibrium, the spring has stretched a distance of x = 17.6 cm, as shown in the diagram. This toy is so adorable you pull the monkey down an additional d = 7.6 cm from equilibrium and release it from rest, and smile with delight as it bounces playfully up and down. 1. Calculate the speed of the monkey, ve, in meters per second, as it passes through equilibrium. 2. Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms of m, x, d, k, the maximum height above the bottom of the motion, hmax, and the variables available in the palette. 3. Calculate the maximum displacement, h, in centimeters, above the equilibrium position, that the monkey reaches.
A child’s toy consists of a…
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U=U(TV). The volume dependence comes from the potential energy due to the
interactions among the particles. For free particles, there are no interactions, thus, the
internal energy U should be independent of the volume, i.e.,
= 0. Verify this
result for ideal gas pV = RT.
au
av
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