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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pdf

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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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