Copy of Lab 4 _ The Human Arm
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1. What is the mass of the arm and an estimate of its uncertainty? Explain how you
estimated the uncertainty.
Angle ( )
◦
M
l
(kg)
M
t
(kg)
M
T
(kg)
δ
90
0
1.885
0.01
50
0
2.200
0.01
50
100
3.970
0.02
130
0
2.050
0.02
Table 4.1:
The table shows the values for the angle of the arm when lifted at a certain angle, the
mass of the load (M
l
), mass applied (M
t
), and the uncertainty of the mass applied ( M
T
).
δ
Using equation 11.4 m
T
d
1
sin(θ) - m
L
(d
1
+ d
2
+ d
3
) - m
A
(d
1
+ d
2
) = 0 where m
T
is the mass applied
(kg), m
l
is the mass of the load (kg), θ is the angle of the metal arm (
°),
and d
1
, d
2
, d
3
are the
given distances of the arm given in the lab manual. By isolating for m
A
, the mass of the arm is
obtained:
m
A
(1.885)(2.35𝑠𝑖?(90))
14.5
=
m
A
= 0.3055 kg
We calculate the value of the mass applied by taking the average of the maximum and minimum
values:
. We then obtain its uncertainty by taking the difference
1.885 + 1.895
2
=
1. 890 𝑘𝑔
between the two m
T
values and averaging them:
For that reason, m
T
1.895 − 1.885
2
= ± 0. 005.
=
. Accordingly, the uncertainty of the mass of the arm (m
A
) is:
(1. 890 ± 0. 005)
.
(0. 3055 ± 0. 005)
2. Using Eq. 11.4, what is the theoretical value of
𝑚
T
(no uncertainty required) that will
balance the arm for
𝜃
= 50
°
? Show your work. What is the experimental value (estimate its
uncertainty) and is it consistent with the theoretical value?
m
T
d
1
sin(θ) - m
L
(d
1
+ d
2
+ d
3
) - m
A
(d
1
+ d
2
) = 0
m
T
=
?𝐴(?1 + ?2) ?1𝑠𝑖?(50)
= 0.3055 * 14.5
2.35*𝑠𝑖?(50)
=
2. 460 𝑘𝑔
When isolating for m
T
, the mass applied is neglected because there was no mass attached to the
arm. Therefore, the theoretical mass value of m
T
that will balance the arm for
𝜃
= 50
°
is 2.460
kg. The experimental value for the mass applied is found to be 2.200 kg and its uncertainty is
obtained by averaging the difference between its maximum and minimum values:
Its uncertainty is then obtained by averaging the difference between
2.200 + 2.210
2
=
2. 205 𝑘𝑔.
the maximum and minimum values:
As a result, the experimental
2.210 − 2.200
2
= ± 0. 005.
value for m
T
is
Comparing the experimental and theoretical values of m
T
, it
(2. 205 ± 0. 005).
can be observed that there is a moderate agreement between them.The difference between the
values (2.460-2.205 = 0.255) which falls within the standard deviations of the theoretical value
proving there is an agreement between the values. Hence, the experimental value of m
T
is
consistent with its theoretical value with no mass load applied.
3. Show your work for calculating the force at the elbow joint (
??
) for
𝜃
= 50
°
and
𝑚𝐿
= 0
g. Do not include uncertainty.
To calculate the force at the elbow joint, equation 11.5 must be used:
F
x
= 0 and
F
y
= 0.
Σ
Σ
𝑇??𝑠(θ) = 𝐹???𝑠(θ)
gm
T
cos( ) = F
E
cos(
θ
ϕ)
1.889*9.81*cos(50
°
= F
Ex
)
F
Ex
= 11.90 N, to the left
F
y
= 0
-sin( )m
T
g - m
A
g - m
L
g - F
E
sin( ) = 0
θ
θ
F
Ey
= [-sin(50)*
2.2
9*9.81] -[ 0.3055*9.81 + 0*9.81]
F
Ey
= 16.50 N, downwards
Thus, F
E
=
𝐹?
2
+ 𝐹?
2
= (− 11. 90)
2
+ (− 16. 50)
2
= 20. 34𝑁. 4. Calculate and measure
𝑚𝑇
for 50
°
and a load (either 100 g or 150 g depending on your
setup). Calculate
??
(no uncertainty). You should include an estimate of the uncertainty for
the measured
𝑚𝑇
. How do these values compare to the answers without a load? Show your
work for all parts of this question.
Using equation 11.4, the theoretical mass applied can be isolated from the equation as follows:
m
T
=
?𝐿(?1+?2+?3) + ?𝐴(?1 + ?2) ?1𝑠𝑖?(50)
=
[0.1*0.29]+[0.3055*0.145]
0.0235*𝑠𝑖?(50)
=
4. 072 𝑘𝑔 However, the experimental mass obtained from the lab experiment is 3.970 kg. To calculate the
experimental value of m
T
, the average mass values are taken:
The
3.970+3.990
2
=
3. 980 𝑘𝑔.
uncertainty is then found by:
Hence, the experimental value of m
T
when
3.990−3.970
2
= ± 0. 01.
m
L
is 150 grams is
Comparing the theoretical and experimental values of m
T
,
(3. 980 ± 0. 01).
it is observed that the difference between the values (4.072 - 3.980 = 0.092) and the uncertainty
is (0.01 - 0 = 0.01), which does not fall within the standard deviation of the theoretical value,
showing that there is a slightly poor agreement between them. The experimental value of m
T
is
not consistent with its theoretical value due to the human error of miscalculating the weights
applied and not properly balancing the arm when the apparatus is at 50° with a load of 0.100 kg
applied.
To calculate the force at the elbow joint (F
E
), equation 11.5 is used.
F
x
= 0;
F
x
= m
T
gcos( ).
Σ
Σ
θ
Using this, F
Ex
= 4.072*9.81*cos(50) = 25.68N. For the force in the y-direction,
F
y
= 0;
F
y
=
Σ
Σ
m
T
gsin( ) - m
A
g - m
L
g. Therefore, F
Ey
= [4.072*9.81*sin(50)] - [0.3055*9.81] - [0.100*9.81] =
θ
26.62 N. F
E
is then calculated by:
𝐹 = 𝐹?
2
+ 𝐹?
2
= (25. 68)
2
+ (26. 62)
2
=
36. 99𝑁.
The force F
E
measured without a load (m
L
= 0kg) registers at 20.34 N, while with a load (m
L
=
0.100kg) and the arm at a 50° angle, it rises to 36.99 N, indicating a difference of 16.65 N
between the two. The force F
E
with m
L
= 0.100 kg is 1.819 (36.99/20.34 = 1.819) times greater
than F
E
with mL= 0 kg. Additionally, the mass m
T
recorded without a load is 2.460 kg, whereas
with the 0.100 kg load, it increases to 4.072 kg, marking a difference of 1.612 kg. Consequently,
m
T
with the 0.100 kg load is 1.655 times greater than mT with 0 kg load. Thus, it is evident that
the 0.100 kg load affects both m
T
and F
E
at 50°, increasing them by approximately the same
factor of 1.655 and 1.819, respectively.
5. Calculate and measure
𝑚𝑇
for 130◦ and no weight. You should include an estimate of the
uncertainty for the measured
𝑚𝑇
. Show your work for all parts of this question.
To calculate m
T
when the angle is 130°, equation 11.4 (m
T
d
1
sin(θ) - m
L
(d
1
+ d
2
+ d
3
) - m
A
(d
1
+ d
2
)
= 0) must be used to isolate for m
T
:
m
T
=
?𝐴(?1 + ?2) ?1𝑠𝑖?(50)
= 0.3055 * 14.5
2.35*𝑠𝑖?(130)
=
2. 460 𝑘𝑔
2.050+2.070
2
= 2. 060𝑘𝑔,
2.070−2.050
2
= ± 0. 01 ⇒ (2. 060 𝑘𝑔 ± 0. 01)
m
T
=
(2. 060 𝑘𝑔 ± 0. 01)
6. Do you notice any difference in the motion of the arm at 50◦ vs. 130◦ ? If so, what?
When exploring human arm mechanics, notable disparities arise in its behavior at different
angular positions, specifically at 50° and 130°. When the arm extends to 50° from its starting
point, it retracts to its original position in an anticlockwise motion, aligning with the forward
motion initiated by my lab partner. However, when extended to 130°, the arm retraces its path in
a clockwise manner, seemingly reversing the direction of the force applied. Furthermore, despite
both 50° and 130° positions resulting in identical values of F
E
(20.34 N), the arm's stability
contrasts markedly between the two angles. At 130°, maintaining stability becomes more
challenging as the arm consistently swiftly returns to its initial position, displaying heightened
sensitivity to rapid movements and masses. In contrast, at 50°, the arm exhibits a greater capacity
to sustain its horizontal position, demonstrating more stability.
7. Does this simplified system seem like a good approximation of a human arm? Why or
why not? If you had additional resources, how might you modify the apparatus to be more
like a human arm?
While this simplified model of the human arm isn't perfect, it serves its purpose reasonably well.
It offers a clear visual representation of the arm's angles during movement and aids in
understanding how different parts of the arm function, especially when lifting objects of varying
weights. To enhance the apparatus, I propose adding a hand with fingers to explore how gripping
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objects tightly affects arm movement. Furthermore, attaching a sensor capable of detecting
external factors like temperature would provide insights into its impact on the motion of an arm.
I would also incorporate a weight at the bottom of the arm to lower its center of gravity, thereby
increasing stability and reducing sensitivity to rapid movements caused by different masses,
particularly at 130°.
8. Acknowledgements & References
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Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Engineering Mechanics: Statics
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ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY