Statics in the human body

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Temple University *

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1061

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Mechanical Engineering

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Apr 3, 2024

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T E M P L E U N I V E R S I T Y P H Y S I C S 3/14/2024 1:31 PM 1 Statics in the human body The human body is truly an amazing machine, and it can be understood, in part, using the concept of static equilibrium. An object in static equilibrium isn’t accelerating or rotating, which means that all the torques and forces acting on the object must sum to zero: ∑ ࠵?⃗ = 0 ∑ ࠵? ⃑ = 0 In principle, a given force could produce both translation and rotation. The torque produced by a force ࠵? is defined in vector form by the equation ࠵?⃗ = ࠵?⃗ × ࠵? ⃗ . In this lab we’ll concern ourselves with only the magnitude of the torque, thus simplifying the equation to ࠵? = ࠵?࠵? sin θ . (Equation 1) The diagram below defines the terms in Equation 1. The lever arm ࠵? is the line connecting the pivot and the point of application of the force. The angle θ is the angle between the force and the lever arm. Learning goals for this lab: • Understand the conditions for static equilibrium when both force and torque are acting. • Show how joints in the body can be modeled by simple levers. • Derive the relations showing mechanical advantages and disadvantages of levers. Apparatus: pivot stand, meterstick, pivot bolt and wingnut, 2 sliding mass hangers, 500 g mass, wireless force sensor. Part I. Standing on tiptoes: the class II lever A class II lever is one where the load and the applied force are on the same side of the fulcrum, but the load is closer to the fulcrum than the applied force. Two examples of the class II lever are the foot standing on tiptoes and the wheelbarrow, as shown in Figure 1 below. θ ࠵? ࠵?

T E M P L E U N I V E R S I T Y P H Y S I C S 3/14/2024 1:31 PM 2 a. Set up your apparatus as in the photo above. To do this, turn on the force sensor and open the Capstone software. In the hardware setup menu, click on your force sensor (if multiple sensors appear, identify yours using the ID no. on the sensor). From the display option menu at right, make a digits display of the force as well as a graph of force vs. time. Click Record to see live data, then without any force acting on it, zero the force sensor using the zero button at the bottom of the screen that looks like this: b. Weigh and record the mass of the meterstick for later (the mass will be needed in Part III). Then mount the meterstick in the stand at the fifth hole from the bottom as in the photo above by slotting the meterstick into the stand then passing the pivot bolt through the holes in the stand and the 1 cm mark of the meterstick. Secure the pivot bolt using the wingnut provided. c. Slide the blue mass hangers onto the meterstick as in the photo above, making sure the second hanger is inverted so the force sensor can be hooked onto it. Hook the 500 g weight on the first hanger (the total weight of the load is 510 g since the mass hanger is 10 g). pivot force sensor mass hangers 500 g mass

T E M P L E U N I V E R S I T Y P H Y S I C S 3/14/2024 1:31 PM 3 d. Carry out a quick observation on how the location of the load affects the force that must be applied to maintain equilibrium (the precise values of force and position are not important). To do this, apply force by pulling up on the force sensor as in the photo below until the 510 g load is lifted a few millimeters off the benchtop and the meterstick is horizontal. Keep the meterstick horizontal while you vary the lever arm of the 510 g load by sliding it back and forth as shown below. Observe how the force you apply must change for short load lever arms as compared to long load lever arms. For the data section of your report, record your observation as to how the applied force must change as you move the load farther from the pivot. Question 1. In the observation you just made, the meterstick was in static equilibrium. Thus, there was zero net torque on the meterstick and the counterclockwise torque was equal to the clockwise torque. Sketch a free-body diagram that shows the pivot and the meterstick and label the applied force ࠵? A and lever arm ࠵? A as well as the load force ࠵? L and its lever arm ࠵? L . Then, starting with the torque equilibrium equation below, derive an expression for the applied force ࠵? A in terms of the lever arms ࠵? A and ࠵? L and the weight of the load ࠵? L . Assume that θ A and θ L equal 90 degrees. ∑ ࠵?⃗ = ࠵? A − ࠵? L = ࠵? A ࠵? A sin θ A − ࠵? L ࠵? L sin θ L = 0 (Equation 2) Question 2. Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage , by amplifying the input force to provide a greater output force. Use your derived expression from Question 1 to show that the applied force needed will always be less than the weight of the load for class II levers. e. In your experiment, where should the two forces be located to obtain the greatest possible mechanical advantage? Slide the forces to the locations where you obtain the greatest mechanical advantage and apply force via the force sensor. Record the values for ࠵? A and ࠵? L noting that the pivot is at the 1 cm mark so you must subtract 1 cm from the meterstick readings. Use your values of ࠵? A and ࠵? L to calculate the mechanical advantage as a multiplicative factor. Record the values for your lab report. A final note on Part I. As you saw, the lever amplifies our input force, but we know from energy conservation that the energy input must equal the energy output. So, what is exchanged in order to obtain the force increase? Displacement. When you lift the load, you can see that the vertical displacement of your hand is much greater than the vertical displacement of the load. Vary the load’s lever arm by sliding it along the meterstick.

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T E M P L E U N I V E R S I T Y P H Y S I C S 3/14/2024 1:31 PM 4 Part II. The bicep curl: a class III lever Class III levers are the most common type in the human body; one example being the biceps when it is flexing to lift the forearm. As with the class II lever, the weight and muscle are on the same side of the fulcrum, but now the muscle is closer to the fulcrum. Question 3. For Part II, let’s call the applied force ࠵? muscle and the weight of the load ࠵? weight and their respective lever arms ࠵? muscle and ࠵? weight as in the figure above, left. Set up the equilibrium equation just as in Part I by setting the sum of the torques to zero. Solve for ࠵? muscle in terms of ࠵? muscle , ࠵? weight , and ࠵? weight . What equation did you obtain for ࠵? muscle ? Is it the same as equation 2 (other than the subscripts)? Make a prediction using your equation: if ࠵? muscle = 5 cm and ࠵? weight = 30 cm, how much force does the muscle need to generate to lift the 510 g? You can ignore the mass of the meterstick and the mass of the hanger attached to the force sensor. a. Now do a quick check of your prediction by setting up your apparatus to mimic the biceps (you’ll need to swap the position of that mass hangers from Part I so that the applied force is closer to the pivot than the load). Zero the force sensor and hook it to the mass hanger and slide the mass hanger to the 6 cm mark on the meterstick (this way ࠵? muscle = 5 cm). Place the load hanger at the 31 cm mark (this way ࠵? weight = 30 cm) as in the photo above. In the same manner as Part I of the lab, pull upward on the force sensor until the meterstick is horizontal and the load is just off the table. You’ll have to pull hard! You may need a lab mate to hold down the platform so it doesn’t lift off the table. Record your values of ࠵? muscle , ࠵? muscle , ࠵? weight , and ࠵? weight in a data table for your lab report. Question 4. Up to now we’ve only considered torque. However, the net force is also zero for an object in equilibrium. Compare the value you obtained for ࠵? weight and ࠵? muscle : are they equal? Does this mean there must be another force acting? Which direction is it pointing given that ࠵? weight is pointing downward and ࠵? muscle is pointing upward? Considering the fact that this third force does not produce torque, where must

T E M P L E U N I V E R S I T Y P H Y S I C S 3/14/2024 1:31 PM 5 it be acting? Note that in the human arm, this third force is a compressive reaction force and can cause injury to the shoulder joint when attempting to lift weights that are too large. Part III. Avoiding back injuries Let’s move to a more complex situation: the torso. We’ll model the torso rotating around the hips as a class III lever just like we did for the forearm above. However, there is one big difference: the muscles (erector spinae in this case) do not pull upward the way the bicep muscle does, instead they pull nearly parallel to the torso. We’ll again use static equilibrium to show why back muscles are easily overtaxed when one is trying to rotate their torso to the upright position using the hips as the pivot. a. On your apparatus, set up the arrangement as in the image above. To do this, attach the large metal hook through the set of holes three places above the pivot and secure it using the wingnut provided. Then slide the force sensor hanger to the 26 cm mark and fix it in place with its screw. Next lift the meterstick upward until you can hook the handle side of the force sensor to the large hook so the force sensor is suspended between the vertical beam and meterstick as shown above. Lastly, secure the 510 g load to the 51 cm mark on the meterstick. Now the forces will act at angles that are no longer 90 degrees. b. In a data table, record ࠵? muscle (the force sensor reading), ࠵? weight (the 510 g load plus the weight of the meterstick), and the lever arms ࠵? muscle (25 cm) and ࠵? weight (50 cm) as well as their angles θ muscle and θ weight . The angles can be read from the mass hangers’ built-in compass (you may need your phone light to help you see the small print angle numbers). Important: since the hanger angles are with respect to the normal, you must subtract the value from 90 degrees so it will be with respect to the beam instead of the normal. As a check, typical values for the angles are about ࠵? muscle = 25 ° and ࠵? weight = 65 ° . c. Draw a free-body diagram for this arrangement that includes the muscle force ࠵? muscle , the force of the weight ࠵? weight , and their lever arms ࠵? muscle and ࠵? weight as well as their angles θ muscle and θ weight . Be sure to clearly label both forces, their lever arms, and their angles.

T E M P L E U N I V E R S I T Y P H Y S I C S 3/14/2024 1:31 PM 6 d. With your free-body diagram to help you, use torque equilibrium to write an expression for the net torque on the meterstick. Then solve the expression to obtain an expression for the muscle force in terms of the other quantities. Check the equation with your instructor. e. Use your expression for the muscle for to estimate the theoretical force required of the muscle to maintain static equilibrium by plugging in your values for ࠵? weight , ࠵? muscle , ࠵? weight , θ muscle , and θ weight . Question 5. In a real human body, the value of θ muscle is probably closer to 10 degrees. How is the magnitude of the torque affected when the angle becomes very small? (Try plugging sin 10 degrees into your calculator to see the value.) Use this to explain why the erector spinae muscles must produce very large forces to keep the torso in static equilibrium when leaning over. Question 6. Conversely, to avoid overtaxing the erector spinae muscles one should keep the torso more upright when lifting objects. What happens to the magnitude of the torque produced by the weight when the torso is vertical? Hint: what’s the value of θ weight when the torso is vertical? Use this to explain why lifting heavy objects with your torso vertical helps to avoid injury to the erector spinae muscles.

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