EP Lab 8

pdf

School

Temple University *

*We aren’t endorsed by this school

Course

1061

Subject

Mechanical Engineering

Date

Apr 3, 2024

Type

pdf

Pages

Uploaded by SargentDugong7894

Lab 7: Statics in the Human Body Group members: Angelina Talucci, Lilly Drasin, Samayra Arce Goals: ● Understand the conditions for static equilibrium when both force and torque are acting. ● Show how simple levers can model joints in the body. ● Derive the relations showing the mechanical advantages and disadvantages of levers. Procedure: Part 1: 1. Set up the apparatus as shown in the picture below. To do this, turn on the force sensor and open Capstone. In the hardware setup menu, click on the force sensor, create a digit display of the force, and graph of force vs. time. Click record to see live data, and without any force acting on the sensor, zero the force using the zero button. 2. 3. Weigh and record the mass of the meter stick (needed in part 3), and mount the meterstick as shown in the picture. 4. Slide the blue mass hangers onto the meter stick, ensuring the second hanger is inverted so the force sensor can be attached. Hook the 500g weight on the first hanger (total weight = 510g) 5. Observe how the location of the load affects the force that must be applied to maintain equilibrium. Apply force by pulling the force sensor up until the 510g load is lifted a few mm off the benchtop, and the meter stick is horizontal. Keep the meter stick horizontal as the hanger is slid back and forth. Observe how the force applied is changed for short-load and long-load levers. 6. Record observation as to how the applied force must change as you move the load farther from the pivot.

7. 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 r A and r 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 r A and r L to calculate the mechanical advantage as a multiplicative factor. Record the values in the results. Part 2: 1. 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 the mass hangers from Part 1 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 meter stick (so r muscle = 5 cm). Place the load hanger at the 31 cm mark (so r weight = 30 cm). In the same manner as part 1, pull upward on the force sensor until the meter stick is horizontal and the load is just off the table. Record values of F muscle , r muscle , r weight and F weight in the results section. Part 3: 1. On the apparatus, set up the arrangement as pictured below. Attach large metal hook through the set of holes three places above the pivot and secure it using the wingnut provided. Then slide the force hanger to the 26 cm mark, and fix it in place with screws. 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 the meterstick. Secure the 510 g load to the 51 cm mark on the meter stick. Now the forces will act at angles that are no longer 90º. 2. In Data table, record F muscle (force sensor reading), Weight (the 510g load plus the weight of the meterstick), and the lever arms r muscle (25 cm) and r weight (50cm) as well as their angles θ muscle and θ weight . These angles can be read from the mass hangers’ built-in compass. Since hanger angles are with respect to the normal, subtracting the value from 90º so it will be with respect to the beam instead of normal.

3. Draw a free-body diagram for this arrangement that includes the muscle force F muscle , the force of the weight F weight , and their lever arms r muscle and r weight as well as their angles θ muscle and θ weight . 4. In addition, use torque equilibrium to write an expression for the next torque on the meter stick. Then solve the expression to obtain an expression for the muscle force in terms of other quantities. 5. Use your expression for the muscle force to estimate the theoretical force required of the muscle to maintain static equilibrium by plugging in all the force, lever and angle values. Errors and Precautions: 1. A precaution is to make sure that the force sensor is zeroed. If not, this will have a major effect on the data results. 2. Another error could be not having the apparatus set up correctly. This will have a reflection on all the force data. 3. Another precaution is to make sure that the masses measured are correct and accurate. Say the meterstick weight is inaccurate, this will have a reflection on the results in part 3. Results: Part 1: Observation from the movement of weight: When the force of the load is closer to the force sensor, more applied force must be present to keep the angle at 90º. Short-arm levers require more force to maintain equilibrium than long-arm levers. Free-Body Diagram:

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Expression (from Question 1): F A = (r L F L sin 𝜃 L ) / (r A sin 𝜃 A ) Table: F A (applied force) 4.1 N F L (load force) 7.9 N r A (lever arm of F A ) 51 cm = 0.51 m r L (lever arm of F L ) 26 cm = 0.26 m 𝜃 A 90º 𝜃 B 90º Part 2: F muscle (applied force) 2.8 N F weight (load force) 4.998 N r muscle (lever arm of F muscle ) 5 cm = 0.05 m r weight (lever arm of F weight ) 30 cm = 0.3 m

Part 3: Free-Body Diagram: Expression of meterstick (from torque equilibrium): T = rFsin 𝜃 meterstick Expression for muscle force: T = rFsin 𝜃 muscle F = T / (r muscle sin 𝜃 muscle ) Estimated theoretical force (plug-in table values): F = 0.506 N (0.25m * sin(25) = 0.053 N Table: F muscle (force sensor reading) 25.1 N F weight (510g load + meterstick) 567 g = 0.567 kg r muscle (lever arm of F muscle ) 25 cm = 0.25 m r weight (lever arm of F weight ) 50 cm = 0.5 m 𝜃 muscle 25º 𝜃 weight 65º

Questions: 1. In the observation you just made, the meter stick 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 𝜃 A and 𝜃 L = 90º. ∑ 𝜏 = 𝜏 A - 𝜏 L = r A F A sin 𝜃 A - r L F L sin 𝜃 L = 0 r A F A sin 𝜃 A = r L F L sin 𝜃 L F A = (r L F L sin 𝜃 L ) / (r A sin 𝜃 A ) 2. Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force to provide greater output force. Use your derived expression from Q1 to show what applied force needed will always be less than the weight of the load for class II levers. r A F A sin 𝜃 A = r L F L sin 𝜃 L MA = F L / F A = 1/Sin 𝜃 MA = (r A F A sin 𝜃 / r L ) / (r L F L sin 𝜃 / r A ) MA = 1 / sin 𝜃 3. For part 2, lets 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 shown in the figure below. Set up the equilibrium equation just as in Part 1 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 was obtained for F muscle ? Is it the same as equation 2 (other than 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? Ignore the mass of the meterstick and the mass of the hanger attached to the force sensor. r muscle F muscle sin 𝜃 muscle - r weight F weight sin 𝜃 weight = 0 r muscle F muscle sin 𝜃 muscle = r weight F weight sin 𝜃 weight F muscle = r weight F weight sin 𝜃 weight / (r muscle sin 𝜃 muscle ) F muscle = (0.3 m)(0.510kg)(sin65º) / (0.05)(sin25) F muscle = 6.56 N

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

If we predict using this equation with r muscle = 5 cm and r weight = 30 cm, we can estimate how much force the muscle needs to generate to lift the 510g load. By plugging these values into the equation, we can find the force required. 4. Up to now we’ve only compared torque. However, the net force is also zero for an object in equilibrium. Compare the value you've 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? Because 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 too large. Comparing the forces Fweight (acting downward due to the load's weight) and Fmuscle (acting upward due to muscle exertion) reveals they are not equal in magnitude, but the object is in equilibrium. This means that another force is required. This third force counteracts the combined effects of Fweight and Fmuscle. Acting opposite to their resultant force, it would point downward and act directly at the pivot point, providing support without contributing to torque. 5. In a real human body, the value of 𝜃 muscle is probably closer to 10º. How is the magnitude of the torque affected when the angle becomes very small? (Try plugging sin(10) into a calculator to see the value). Use this to explain why the erector spinal muscles must produce very large forces to keep the torso in static equilibrium when leaning over. The magnitude of torque (τ) is given by the formula: τ=F×r×sin (θ). When the angle θmuscle is very small, sin (θ)gets very close to 0. Therefore, the torque equation becomes: τ=F×r×0. When the angle between muscles and the body's pivot point is very small, the muscles lose their leverage, making it harder to maintain balance. Since sin θmuscle approaches zero when θmuscle is small, the torque produced by these muscles becomes very small. The erector spinae muscles

must produce very large forces to counteract this small torque and maintain static equilibrium. Essentially, they need to compensate for the reduced lever arm effectiveness by generating much larger forces to counteract the gravitational torque pulling the torso downward. 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 vertically helps avoid injury to the erector muscles. This is because the angle of the torso is 0, and since T = rFsin 𝜃 , sin(0) = 0, this means that no torque is exerted by the weight. Keeping the torso vertical will reduce the amount of force and torque that is applied to the spinae muscles, reflecting on reducing injury. Conclusion: In conclusion, our data revealed that as the load is positioned closer to the force sensor, a greater applied force is required to maintain equilibrium. This can be proven from torque equilibrium with the relationship between the applied force (F A ), lever arms (r A and r L ), and the weight of the load (F L ). This correlates to the mechanical advantage inherent in levers. Specifically, the equation F A = (r L F L sin 𝜃 L ) / (r A sin 𝜃 A ) gave a relationship between the applied force and how it’s influenced by the lever arms and the load's weight. Analysis of net forces highlighted the need for a support force acting at the pivot point, ensuring equilibrium without contributing to torque. This was underscored by the equation r muscle F muscle sin 𝜃 muscle - r weight F weight sin 𝜃 weight = 0 which showed the balance of forces required for stability.