Copy of Lab 8 Statics in 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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Lab 7 Concussions Goals - The goal of this experiment is to understand standard equilibrium when force and torque are acting, show how joints in the body can be modeled with levers, and to show the mechanical advantages and disadvantages of levers. Procedure - For part I, we observed how the location of the load affects the force that must be applied to maintain equilibrium. We did this by setting up our apparatus, weighing the mass of the meterstick, and mounting the meter stick in the stand. We then slid the blue mass hanger onto the meterstick, hooked the sensor onto it, and then hooked the 500 g weight on the first hanger. For our observation, we pulled the sensor to apply force to see how the force would change. - For part II, we set up the apparatus to mimic the biceps, pulled upward in the force sensor, and recorded the data. - For part III, we estimated the theoretical force required of the muscle to maintain static. We did this by setting up the apparatus so that the forces are acting at angles and recording the force sensor reading, 510 g load plus the weight of the meterstick, the 25 cm of the sensor hanger, the 50 cm of the 510 g load, and their angles. We then drew a free-body diagram that included everything listed. We used the diagram to find the expression for the net torque on the meterstick and were able to estimate the theoretical force required. Error and Precautions - A possible error is friction. Friction could cause the hanger to remain stationary - Another possible error could include the setup, if we didn’t put the meter stick or hangers in the correct places, it could have caused inaccurate data. Results - In part I, when the weight was closer to the stand, the force was lower. When the weight was moved closer to the force sensor, the force increased. - F muscle r muscle F weight r weight 34.95N 5cm 510g 30cm -

- Questions 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 𝐹 A and lever arm 𝑟 A as well as the load force 𝐹 L and its lever arm 𝑟 L . Then, starting with the torque equilibrium equation below, derive an expression for the applied force 𝐹 A in terms of the lever arms 𝑟 A and 𝑟 L and the weight of the load 𝐹 L . Assume that θA and θL equal 90 degrees. - - 𝐹 𝐴 = ? 𝐿 𝐹 𝐿 ? 𝐴 Question 2. Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, 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. - ? 𝐴 = 49, ? 𝐿 = 24 - 𝐹 𝐴 = 24(510) 49 ≈ 250 < 510 Question 3. For Part II, let’s call the applied force 𝐹 muscle and the weight of the load 𝐹 weight and their respective lever arms 𝑟 muscle and 𝑟 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 𝐹 muscle in terms of 𝑟 muscle , 𝑟 weight , and 𝐹 weight . What equation did you obtain for 𝐹 muscle ? Is it the same as equation 2 (other than the subscripts)? Make a prediction using your equation: if 𝑟 muscle = 5 cm and 𝑟 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. - , , τ ? = 𝐹 ? × ? ? τ 𝑤 = 𝐹 𝑤 × ? 𝑤 τ ? + τ 𝑤 = 0 𝐹 ? × ? ? + 𝐹 𝑤 × ? 𝑤 = 0 𝐹 ? =− 𝐹 𝑊 ×? 𝑤 ? ? - 𝐹 ? = (−0.510?𝑔)×(9.8?/? 2 )×(0.3?) (0.05?) = 15. 348𝑁

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 𝐹 weight and 𝐹 muscle : are they equal? Does this mean there must be another force acting? Which direction is it pointing given that 𝐹 weight is pointing downward and 𝐹 muscle is pointing upward? Considering the fact that this third force does not produce torque, where must 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. - For a state of static equilibrium, an object, arm in this case, should be at rest, where it should have no translation motion, nor should it have rotational motion. Thus there should be no net force acting. The net torque and net force on your arm are both zero. Question 5. In a real human body, the value of θmuscle 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. - This shows that the magnitude of the torque decreases as the angle becomes ?𝑖?(10 ◦ ) = 0. 17 smaller. Because sin(10 degrees) is so small and causes a less effective torque, spinal muscles must produce large forces to combat the less effective torque to keep the torso in static equilibrium. 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 θweight when the torso is vertical? Use this to explain why lifting heavy objects with your torso vertical helps to avoid injury to the erector spinal muscles. - When the torso is kept more upright while lifting objects, the angle θ weight between the weight force and the lever arm decreases. In other words, θ weight approaches 0 degrees when the torso is vertical. This results in a decrease in the magnitude of the torque produced by the weight. The torque produced by a force at an angle from the lever arm is given by the formula: . When θ weight is close to 0 degrees (torso vertical), sin(θ weight ) approaches 0. τ = 𝐹 × ? × ?𝑖?(θ) Therefore, the torque produced by the weight decreases significantly. Lifting heavy objects with the torso more upright helps avoid overtaxing the erector spinae muscles because the reduced torque means there is less rotational force exerted on the spine. This reduces the stress and strain on the erector spinae muscles, which are responsible for maintaining the upright position of the spine during lifting tasks. Discussion - We predicted that when 510 g was closer to the sensor, more force was needed, but when it was closer to the stand, less force was needed. This was seen because the applied force from part I came out to about 250 N, which is less than 510 g. For part II, we predicted that the force needed for a muscle to lift 510 g would be 15.348N and the results came to about 34.95N. For part III, we predicted the force needed for a muscle to keep the torso at static equilibrium would be high. For the results, we found that the force needed was about 26.55 N with . θ ?????? = 45??𝑔 𝑎?? θ 𝑤?𝑖𝑔ℎ? = 70??𝑔

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