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©UDel Physics 1 of 8 Spring 2019
PHSY201 Lab 9 – Biomechanics of
Torques and Forces in Equilibrium
Instructional Goals:
Goals:
 Learn how to apply torques and forces to equilibrium situations.
 Get a feeling for how the material we have learned so far can be applied to biomechanics.
 Gain more experience comparing measured values.
What You Turn In:
 Pre‐Lab in the first five minutes of your lab section.
 A Full Lab Report according to the instructions provided in the Lab Report Guidelines.
Background Reading:
Relevant background can be found in the lecture notes and Chapter 11 of Walker 4th edition.
Equipment provided by the lab:
 Metal poles: 1 long, 1 medium
 Table clamp (or table post mount)
 Wireless Force Sensor
 Rotating clamp
 Ring Clamps (4)
 Hanging mass holder with 50g and 100 g masses
 Metal chain (or inextensible string)
 Meter stick and/or tape measure
 Level
Equipment provided by the student:
 Ruler
 Pen
 Calculator
 USB memory device for taking computerized data from the lab (or alternatively email the data to
yourself
PHYS201 Lab 9 Torques and Forces in Static Equilibrium
©UDel Physics 2 of 8 Spring 2019
Background:
Torque and Rotation
Consider a mass m attached to a rotation axis by a rigid but
massless rod of length r as shown to the right.
Suppose a force F

is applied to the mass. The only component
of the applied force that can cause rotation is that which is
perpendicular to both the rotation axis and the rigid, but
massless, rod. Denote the magnitude of this component as F .
In the plane perpendicular to the rotation axis, the setup (with
only the component of force F shown) is
Where from the drawing above, it is clear that sin
F F 
  .
From Newton’s second law,
F ma
 

Multiplying both sides by the distance from the axis of rotation at which the force acts,
rF mra
 

Since at a r
  , we have for this mass,
2
rF mr I
 
  
The quantity, rF , is known as the torque caused by the applied force F

. Thus, we can define the left side of the
equation as sin
rF
 
 , or in vector notation,  
τ r F



, where it the magnitude is given as before. On the
right‐side we have the moment of inertia, I, which is
2
I mr
 . This is a quantity that depends on the distribution
of mass in an object. That is the resulting rotation and changes in rotation are dependent on where each little bit
of mass is located in the system relative to the axis of rotation (or pivot).
Thus, Newton’s second law of motion, in a rotational sense for a point of mass is I
 
 .
We can recast Newton’s 2nd
law for the case of rotation or an extended object about a pivot as net I
 
 . Where
I is the summed total of all the moments of inertia for the entire object. In practice, it is calculated using calculus
methods. For now, we take it as the rotational analog of mass, just as  is the rotational analog of force and 
is the rotational analog of acceleration. More on this in the next lab.
At equilibrium the angular acceleration, α, is zero, so the net torque must be zero. So, if an object is not rotating
at rest the net torque must be zero, 0
net i
i
 
 
 . That is the sum of all torques about any given pivot point
must be zero. We are free to choose the pivot point as the object is not rotating. In a problem such as this, it is
m
r
F

ϕ
F

ϕ
Overhead view
PHYS201 Lab 9 Torques and Forces in Static Equilibrium
©UDel Physics 3 of 8 Spring 2019
best to pick the most convenient pivot point as possible. I.e., pick a pivot point that reduces the number of
variables you need to calculate/measure.
Recall here that the torque is defined as sin
Fr
 
 and we are confining ourselves to two‐dimensional motion
only.
Where F is the force applied, r is the distance from the pivot to the applied force and θ is the angle between the
vectors F

and r

. By convention, we take the algebraic sign of the torque as positive for torques that would cause
a counterclockwise (CCW) rotation in the absence of any additional torques. It is negative for torques that would
cause a clockwise (CW) rotation. This is the essence of the right‐hand rule.
Center of Gravity: For an extended object, such as the boom, we can think of the weight acting at one point, the
center of mass, for the purposes of calculating the torque. This is the point where the extended object would
balance. In this lab, you will find the center of gravity of a boom by balancing it on a meter stick’s edge and
marking that place on the boom with a pencil or chalk (something erasable). Since the boom will have some
attached equipment (e.g. ring hangers), its center of mass will not be at is volumetric center.
Besides the angular acceleration being zero, an additional condition for equilibrium is that the linear acceleration
be zero 0

a

. So, we have the additional relations for the components of the vectors (in two dimensions):
0 0
x y
F and F
 
 
which we can use to analyze a system at equilibrium. We will apply these at the end of the lab to find the net force
the elbow is applying.
In this lab, the system we will be examining is an “arm” holding a weight. See the figure below (Figure 1).
B
F

A
W

h
W

A
W

h
W

B
F


A h
B
E
s
Figure 1 ‐ Human Arm Analog. The horizontal beam represents the forearm with a center of mass at A. Any weight in the hand
is at point h and the bicep muscle attaches at point B on the forearm (radius) and point s at the shoulder (scapula). Point E is
the elbow, which experiences a force which will be calculated at the end of the lab.
PHYS201 Lab 9 Torques and Forces in Static Equilibrium
©UDel Physics 4 of 8 Spring 2019
Please note the following from Figure 1:
 The forearm (radius/ulna) in our mock up is a horizontal boom from E (elbow) to hand (h).
 The bicep muscle is a chain, with a force meter attached running from s (shoulder) to B (forearm).
 The elbow is a joint between the vertical metal rod and the horizontal boom. The point where the joint
swivels is E in the diagrams above. Note that in our mock up, the rod does not end at E but extends
past this point. You need to make sure you are measuring from the correct point when getting
distances.
 We have excluded other muscles (e.g. triceps, etc.) which cause other torques on the “bones” in our
system and the attachment points of the biceps are approximated. However, it is quite accurate for our
needs.
In order to calculate our torques, we must set a point about which we would rotate. In this system, we’ll use the
natural choice of the elbow, E. We choose this point because we cannot directly measure the forces applied here
and since sin
Fr
 
 , any force applied at a distance of r = 0, provides no torque. We will calculate the
components and net force on the elbow after we have satisfied that the system is not rotating.
We need to define some distance measurements. We define
them according to Figure 2. Every distance measured from
our defined pivot point to any point on the horizontal boom
is given as r with the appropriate subscript. For example, the
distance from the elbow (E) to the center of mass of the
horizontal boom (A) is A
r . The distance from the elbow (E)
to the shoulder (s) is s
 and the distance from the shoulder
(s) to the horizontal boom through the chain (i.e. muscle) at
point (B) is B
 .
To keep our calculations simple, we will make the forearm
horizontal such that gravity always pulls down at a right
angle to the boom. If the upper arm is vertical, we have a
right triangle SEB such that,
2 2 2
B B s
r
 
  and
2 2
sin s s
b b s
r
  

 
 
Table 1: Determination of Uncertainties for
Measured Values
Every measurement you will make will have an uncertainty
associated with it. We will determine the uncertainties of the
measurements in the same ways as before (see Table 1).
We will take these values and find the propagated uncertainty for
the forces and torques.
You will use the measured values to calculate the force of the
muscle and compare it to the value given by the force meter.
Measurement Uncertainty
Mass
±1 in the last digit of the scale
or the amount of fluctuation.
Distance
±1 in the smallest increment of
the ruler/meter stick or
estimate of how well you can
judge the size due to rounded
edges.
Force meter Student Determined
Figure 2 ‐ Defined distances used in this manual.
A
W

h
W

B
F


A h
B
E
s
PHYS201 Lab 9 Torques and Forces in Static Equilibrium
©UDel Physics 5 of 8 Spring 2019
The practical application of this being that if we show that we can accurately calculate the force needed to hold
up the forearm, we need not slice open an actual person’s forearm and replace the bicep muscle with a force
meter to determine the usual force the muscle can provide. If you then know how much force the muscle should
provide, you can then determine a course of physical therapy or determine how much can be lifted in a hand or
many, many other possibilities.
Experimental Procedure:
A. Setup Capstone to read
the force meter.
1. In “Hardware Setup” add the
Wireless Force Sensor.
2. Drag the “Digits” display to
the screen and set it to Force
(N). You should have at least
two points after the decimal
place.
B. Finding the center of mass
of the boom.
1. Remove the boom from the rotating clamp and remove the mass hanger. Balance the boom arm on a
straight edge (e.g. a meter stick) to find the center of mass and mark it on the boom (e.g. with chalk or
pencil). Please note, you should balance the boom without the attachment hardware, but leave the
slip rings. If at any point you move the slip rings, you will change the location of the center of mass
and must start the experiment again from Part B1.
2. Mass the boom on the digital scale or triple beam balance. Remember, do not move the slip rings,
they are part of the forearm!
C. Assemble the Arm Analog and measure the distances and forces:
1. Attach the boom to the vertical support and allow it to swing freely up and down. Do not tighten the
rotating portion of the elbow clamp, it must swing freely. If it were not, additional torques and forces
would be applied. This would be akin to a person having an arthritic elbow.
2. Record the distance from the pivot (E) to the center of mass (A), A
r . Measure and record the distance
from pivot (E) to the chain attachment (B), B
r . Measure and record the distance from pivot (E) to the
hanger (h), h
r . If you remove the horizontal boom or adjust the elbow clamp in any way you must
start from (B.2), again.
3. Calculate and record the value of A
W . This value should not change through the experiment.
4. Attach the inextensible chain with the Force meter at the shoulder point (s) to the forearm at (B).
Initially do not attach the mass hanger and attachment hardware to the bar at (h). In later trials, you
will attach these items and use the scale to determine the actual mass of the hanger, attachment
hardware, and added masses.
PHYS201 Lab 9 Torques and Forces in Static Equilibrium
©UDel Physics 6 of 8 Spring 2019
5. Calculate and record the value of h
W (zero in your first trial; but again, in later trials, be sure to include
the mass of the hanging holder and attachment hardware when calculating this force). Adjust the
position of the slip ring on the vertical support so that the forearm bar is horizontal. Use a level to
check that the bar is horizontal. But do not let the level rest or hang from the bar as this will add weight
and a torque. This can be accomplished by loosening the thumb screw on the collar and sliding the slip
ring up or down. Note, any time you adjust the slip ring at point (s) you will need to record the new
value of s
 for this trial.
6. Before each measurement, zero the force sensor, with no weight hanging from the hook.
7. Press the record button in Capstone and record the measured value of ,
B meas
F . Be careful to include an
uncertainty.
8. Complete the analysis for calculating ,
B calc
F from the lengths and other forces and compare it
quantitatively to ,
B meas
F .
9. Repeat the procedure (B.5 through B.7) for a total of three different masses in the “hand” (the first of
which was zero mass). Note: every time you change the mass you may need to adjust the
chain/string’s length and position at the “shoulder” and angle to level the “forearm”. You should first
adjust the length of the chain by changing which link is hooked to the spring scale, and then fine tune
the setup by loosening the thumb screw on the attachment collar at point (s) and sliding it up and down.
You should choose three masses to hang such that the reading on the force meter spans the range of
the meter, but does not exceed it (50 N). Your first mass should be 0 kg (no mass hanging from the arm
at all.)
Analysis:
A. Setup and solve the torque equation.
1. Since the system is at rest, it is not accelerating in rotational motion, so there are no net torques,
thus we have
0
 
 .
Write an equation using the torques about the elbow for the horizontal boom. Do not insert any
numerical values!
2. Solve the torque equation for B
F , the force provided by the muscle.
i. Solve the problem algebraically such that you can substitute different values for lengths,
masses, etc. as you do different trials.
3. Substitute your measured masses and lengths to get ,
B calc
F .
PHYS201 Lab 9 Torques and Forces in Static Equilibrium
©UDel Physics 7 of 8 Spring 2019
B. Find the uncertainty in the calculated muscle force.
Rather than calculate each value step by step (which can be done), the following equation is provided:
       
2
2 2 2 2
,
sin g
B B
B calc A A A A h h h h
b
F r
g
F r m m r r m m r
r

    

 
      
 
Note that we take the uncertainty in g and sin to be small.
Use the above formula to find the uncertainty in the calculated muscle force for each trial.
Which term (under the radical) is the most important? Are there any terms so small that they may be
safely ignored?
C. Compare Theory and Measurement Quantitatively.
Compare the calculated value of ,
B calc
F with the measured value read on the scale, ,
B meas
F .
To see if your calculated and measured values agree, find out if
   
2 2
, , , ,
2
B meas B calc B meas B calc
F F F F
 
  
If it is, then we have agreement. If not, double check your work.
D. Calculate the force on the elbow.
Finally, calculate the force on the elbow joint (point “E”) and determine if this is reasonable. Remember
force is a vector.
Make sure you have all the data to include in your lab report and your
analysis makes sense.
Complete the lab report as outlined in the Laboratory Report Instructions.
Make sure you include your names, instructor’s name and the course with lab section number on
your document!
PHYS201 Lab 9 Torques and Forces in Static Equilibrium
©UDel Physics 8 of 8 Spring 2019
Pre Lab Questions
(due at the beginning of your lab session)
Names______________________________________________
Lab Section__________________
1. How will you find the center of mass of the forearm boom?
2. Using Figure 2, write down an equation that balances the torques about the pivot point of the elbow (point
E). Use measured values only (i.e. masses and distances), the force of the muscle, B
F and sin .
3. Would a measured force of  
12.2 0.4 N
 be in agreement with a theoretically calculated force of
 
11.3 0.3 N
 ? Show your work.

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9___Biomechanics_Torques_in_Equilibrium.pdf.pdf

  • 1. ©UDel Physics 1 of 8 Spring 2019 PHSY201 Lab 9 – Biomechanics of Torques and Forces in Equilibrium Instructional Goals: Goals:  Learn how to apply torques and forces to equilibrium situations.  Get a feeling for how the material we have learned so far can be applied to biomechanics.  Gain more experience comparing measured values. What You Turn In:  Pre‐Lab in the first five minutes of your lab section.  A Full Lab Report according to the instructions provided in the Lab Report Guidelines. Background Reading: Relevant background can be found in the lecture notes and Chapter 11 of Walker 4th edition. Equipment provided by the lab:  Metal poles: 1 long, 1 medium  Table clamp (or table post mount)  Wireless Force Sensor  Rotating clamp  Ring Clamps (4)  Hanging mass holder with 50g and 100 g masses  Metal chain (or inextensible string)  Meter stick and/or tape measure  Level Equipment provided by the student:  Ruler  Pen  Calculator  USB memory device for taking computerized data from the lab (or alternatively email the data to yourself
  • 2. PHYS201 Lab 9 Torques and Forces in Static Equilibrium ©UDel Physics 2 of 8 Spring 2019 Background: Torque and Rotation Consider a mass m attached to a rotation axis by a rigid but massless rod of length r as shown to the right. Suppose a force F  is applied to the mass. The only component of the applied force that can cause rotation is that which is perpendicular to both the rotation axis and the rigid, but massless, rod. Denote the magnitude of this component as F . In the plane perpendicular to the rotation axis, the setup (with only the component of force F shown) is Where from the drawing above, it is clear that sin F F    . From Newton’s second law, F ma    Multiplying both sides by the distance from the axis of rotation at which the force acts, rF mra    Since at a r   , we have for this mass, 2 rF mr I      The quantity, rF , is known as the torque caused by the applied force F  . Thus, we can define the left side of the equation as sin rF    , or in vector notation,   τ r F    , where it the magnitude is given as before. On the right‐side we have the moment of inertia, I, which is 2 I mr  . This is a quantity that depends on the distribution of mass in an object. That is the resulting rotation and changes in rotation are dependent on where each little bit of mass is located in the system relative to the axis of rotation (or pivot). Thus, Newton’s second law of motion, in a rotational sense for a point of mass is I    . We can recast Newton’s 2nd law for the case of rotation or an extended object about a pivot as net I    . Where I is the summed total of all the moments of inertia for the entire object. In practice, it is calculated using calculus methods. For now, we take it as the rotational analog of mass, just as  is the rotational analog of force and  is the rotational analog of acceleration. More on this in the next lab. At equilibrium the angular acceleration, α, is zero, so the net torque must be zero. So, if an object is not rotating at rest the net torque must be zero, 0 net i i      . That is the sum of all torques about any given pivot point must be zero. We are free to choose the pivot point as the object is not rotating. In a problem such as this, it is m r F  ϕ F  ϕ Overhead view
  • 3. PHYS201 Lab 9 Torques and Forces in Static Equilibrium ©UDel Physics 3 of 8 Spring 2019 best to pick the most convenient pivot point as possible. I.e., pick a pivot point that reduces the number of variables you need to calculate/measure. Recall here that the torque is defined as sin Fr    and we are confining ourselves to two‐dimensional motion only. Where F is the force applied, r is the distance from the pivot to the applied force and θ is the angle between the vectors F  and r  . By convention, we take the algebraic sign of the torque as positive for torques that would cause a counterclockwise (CCW) rotation in the absence of any additional torques. It is negative for torques that would cause a clockwise (CW) rotation. This is the essence of the right‐hand rule. Center of Gravity: For an extended object, such as the boom, we can think of the weight acting at one point, the center of mass, for the purposes of calculating the torque. This is the point where the extended object would balance. In this lab, you will find the center of gravity of a boom by balancing it on a meter stick’s edge and marking that place on the boom with a pencil or chalk (something erasable). Since the boom will have some attached equipment (e.g. ring hangers), its center of mass will not be at is volumetric center. Besides the angular acceleration being zero, an additional condition for equilibrium is that the linear acceleration be zero 0  a  . So, we have the additional relations for the components of the vectors (in two dimensions): 0 0 x y F and F     which we can use to analyze a system at equilibrium. We will apply these at the end of the lab to find the net force the elbow is applying. In this lab, the system we will be examining is an “arm” holding a weight. See the figure below (Figure 1). B F  A W  h W  A W  h W  B F   A h B E s Figure 1 ‐ Human Arm Analog. The horizontal beam represents the forearm with a center of mass at A. Any weight in the hand is at point h and the bicep muscle attaches at point B on the forearm (radius) and point s at the shoulder (scapula). Point E is the elbow, which experiences a force which will be calculated at the end of the lab.
  • 4. PHYS201 Lab 9 Torques and Forces in Static Equilibrium ©UDel Physics 4 of 8 Spring 2019 Please note the following from Figure 1:  The forearm (radius/ulna) in our mock up is a horizontal boom from E (elbow) to hand (h).  The bicep muscle is a chain, with a force meter attached running from s (shoulder) to B (forearm).  The elbow is a joint between the vertical metal rod and the horizontal boom. The point where the joint swivels is E in the diagrams above. Note that in our mock up, the rod does not end at E but extends past this point. You need to make sure you are measuring from the correct point when getting distances.  We have excluded other muscles (e.g. triceps, etc.) which cause other torques on the “bones” in our system and the attachment points of the biceps are approximated. However, it is quite accurate for our needs. In order to calculate our torques, we must set a point about which we would rotate. In this system, we’ll use the natural choice of the elbow, E. We choose this point because we cannot directly measure the forces applied here and since sin Fr    , any force applied at a distance of r = 0, provides no torque. We will calculate the components and net force on the elbow after we have satisfied that the system is not rotating. We need to define some distance measurements. We define them according to Figure 2. Every distance measured from our defined pivot point to any point on the horizontal boom is given as r with the appropriate subscript. For example, the distance from the elbow (E) to the center of mass of the horizontal boom (A) is A r . The distance from the elbow (E) to the shoulder (s) is s  and the distance from the shoulder (s) to the horizontal boom through the chain (i.e. muscle) at point (B) is B  . To keep our calculations simple, we will make the forearm horizontal such that gravity always pulls down at a right angle to the boom. If the upper arm is vertical, we have a right triangle SEB such that, 2 2 2 B B s r     and 2 2 sin s s b b s r         Table 1: Determination of Uncertainties for Measured Values Every measurement you will make will have an uncertainty associated with it. We will determine the uncertainties of the measurements in the same ways as before (see Table 1). We will take these values and find the propagated uncertainty for the forces and torques. You will use the measured values to calculate the force of the muscle and compare it to the value given by the force meter. Measurement Uncertainty Mass ±1 in the last digit of the scale or the amount of fluctuation. Distance ±1 in the smallest increment of the ruler/meter stick or estimate of how well you can judge the size due to rounded edges. Force meter Student Determined Figure 2 ‐ Defined distances used in this manual. A W  h W  B F   A h B E s
  • 5. PHYS201 Lab 9 Torques and Forces in Static Equilibrium ©UDel Physics 5 of 8 Spring 2019 The practical application of this being that if we show that we can accurately calculate the force needed to hold up the forearm, we need not slice open an actual person’s forearm and replace the bicep muscle with a force meter to determine the usual force the muscle can provide. If you then know how much force the muscle should provide, you can then determine a course of physical therapy or determine how much can be lifted in a hand or many, many other possibilities. Experimental Procedure: A. Setup Capstone to read the force meter. 1. In “Hardware Setup” add the Wireless Force Sensor. 2. Drag the “Digits” display to the screen and set it to Force (N). You should have at least two points after the decimal place. B. Finding the center of mass of the boom. 1. Remove the boom from the rotating clamp and remove the mass hanger. Balance the boom arm on a straight edge (e.g. a meter stick) to find the center of mass and mark it on the boom (e.g. with chalk or pencil). Please note, you should balance the boom without the attachment hardware, but leave the slip rings. If at any point you move the slip rings, you will change the location of the center of mass and must start the experiment again from Part B1. 2. Mass the boom on the digital scale or triple beam balance. Remember, do not move the slip rings, they are part of the forearm! C. Assemble the Arm Analog and measure the distances and forces: 1. Attach the boom to the vertical support and allow it to swing freely up and down. Do not tighten the rotating portion of the elbow clamp, it must swing freely. If it were not, additional torques and forces would be applied. This would be akin to a person having an arthritic elbow. 2. Record the distance from the pivot (E) to the center of mass (A), A r . Measure and record the distance from pivot (E) to the chain attachment (B), B r . Measure and record the distance from pivot (E) to the hanger (h), h r . If you remove the horizontal boom or adjust the elbow clamp in any way you must start from (B.2), again. 3. Calculate and record the value of A W . This value should not change through the experiment. 4. Attach the inextensible chain with the Force meter at the shoulder point (s) to the forearm at (B). Initially do not attach the mass hanger and attachment hardware to the bar at (h). In later trials, you will attach these items and use the scale to determine the actual mass of the hanger, attachment hardware, and added masses.
  • 6. PHYS201 Lab 9 Torques and Forces in Static Equilibrium ©UDel Physics 6 of 8 Spring 2019 5. Calculate and record the value of h W (zero in your first trial; but again, in later trials, be sure to include the mass of the hanging holder and attachment hardware when calculating this force). Adjust the position of the slip ring on the vertical support so that the forearm bar is horizontal. Use a level to check that the bar is horizontal. But do not let the level rest or hang from the bar as this will add weight and a torque. This can be accomplished by loosening the thumb screw on the collar and sliding the slip ring up or down. Note, any time you adjust the slip ring at point (s) you will need to record the new value of s  for this trial. 6. Before each measurement, zero the force sensor, with no weight hanging from the hook. 7. Press the record button in Capstone and record the measured value of , B meas F . Be careful to include an uncertainty. 8. Complete the analysis for calculating , B calc F from the lengths and other forces and compare it quantitatively to , B meas F . 9. Repeat the procedure (B.5 through B.7) for a total of three different masses in the “hand” (the first of which was zero mass). Note: every time you change the mass you may need to adjust the chain/string’s length and position at the “shoulder” and angle to level the “forearm”. You should first adjust the length of the chain by changing which link is hooked to the spring scale, and then fine tune the setup by loosening the thumb screw on the attachment collar at point (s) and sliding it up and down. You should choose three masses to hang such that the reading on the force meter spans the range of the meter, but does not exceed it (50 N). Your first mass should be 0 kg (no mass hanging from the arm at all.) Analysis: A. Setup and solve the torque equation. 1. Since the system is at rest, it is not accelerating in rotational motion, so there are no net torques, thus we have 0    . Write an equation using the torques about the elbow for the horizontal boom. Do not insert any numerical values! 2. Solve the torque equation for B F , the force provided by the muscle. i. Solve the problem algebraically such that you can substitute different values for lengths, masses, etc. as you do different trials. 3. Substitute your measured masses and lengths to get , B calc F .
  • 7. PHYS201 Lab 9 Torques and Forces in Static Equilibrium ©UDel Physics 7 of 8 Spring 2019 B. Find the uncertainty in the calculated muscle force. Rather than calculate each value step by step (which can be done), the following equation is provided:         2 2 2 2 2 , sin g B B B calc A A A A h h h h b F r g F r m m r r m m r r                   Note that we take the uncertainty in g and sin to be small. Use the above formula to find the uncertainty in the calculated muscle force for each trial. Which term (under the radical) is the most important? Are there any terms so small that they may be safely ignored? C. Compare Theory and Measurement Quantitatively. Compare the calculated value of , B calc F with the measured value read on the scale, , B meas F . To see if your calculated and measured values agree, find out if     2 2 , , , , 2 B meas B calc B meas B calc F F F F      If it is, then we have agreement. If not, double check your work. D. Calculate the force on the elbow. Finally, calculate the force on the elbow joint (point “E”) and determine if this is reasonable. Remember force is a vector. Make sure you have all the data to include in your lab report and your analysis makes sense. Complete the lab report as outlined in the Laboratory Report Instructions. Make sure you include your names, instructor’s name and the course with lab section number on your document!
  • 8. PHYS201 Lab 9 Torques and Forces in Static Equilibrium ©UDel Physics 8 of 8 Spring 2019 Pre Lab Questions (due at the beginning of your lab session) Names______________________________________________ Lab Section__________________ 1. How will you find the center of mass of the forearm boom? 2. Using Figure 2, write down an equation that balances the torques about the pivot point of the elbow (point E). Use measured values only (i.e. masses and distances), the force of the muscle, B F and sin . 3. Would a measured force of   12.2 0.4 N  be in agreement with a theoretically calculated force of   11.3 0.3 N  ? Show your work.