Document1 - Recovered - Compatibility Mode

Equilibrium Exp. 8: Equilibrium Using Model of Human Arm Hailey Abell William Wagner Physics 1100L Lab 2 November 2023 ABSTRACT An experiment was performed that involved motion, torque, equilibrium, and force. When Newton’s Law is applied to translational motion, constant linear acceleration requires no linear acceleration and has no force acting on it. Rotational motion does not involve rotational acceleration and has no torque acting on it.

INTRODUCTION A body is in a state of equilibrium when it has a constant linear velocity and a constant rotational velocity. Equilibrium can have situations of uniform motion and no motion. It is known that the result of all forces acting on an object must be zero. The result of all torques around an axis must also be zero. The weight of a mass can be determined using F = mg , which is also the force. The magnitude of the torque is determined using the equation: τ = X F ⊥ , where X is the moment arm or the distance from the spin axis to the applied force. In this experiment, an apparatus that had a similar structure to a human arm was used in this experiment to determine the effect of _____ on ___. The equation for equilibrium of the vertical forces is ∑ F = F b cos θ − F h − F f − F e = 0 , where F b = force of bicep, F h = force of hand, F f = force of forearm, and F e = force of elbow. The equation for equilibrium of the torques is ∑ τ = ¿ τ h + τ f – τ b = 0. To determine F e , use the equation: F e = F b cosθ − F h − F f . The equation for F b is shown below where τ h is the torque of the hand and τ f is the torque of the forearm. F b = τ h + τ f ( X b )( cosθ )

In trial three of the experiment, the mass without any weight added along with a radius change was used. The five sets of fifty calculations were repeated for this part. RESULTS Trial One Percent Difference: Trial Two Percent Difference: Trial Three Percent Difference: No weight on “Hand” Measured F b from 1-a F b = 1.4 N % Difference Calculated F b from 1-b F b = -2.83294 N 50-gram weight on “Hand” Measured F b from 1-c F b = 2.8 N Calculated F b from 1-d F b = -1.99315 N 100-gram weight on “Hand” Measured F b from 1-e

F b = 4.2 N Calculated F b from 1-f F b = -3.71577 Questions: 1. Comment on how well the measured “Bicep” force compares to the calculated “Bicep” force for each of “no weight”, “50-gram weight”, and “100-gram weight”. Include the % difference values for each in your comment. Is there one set that agrees better than the other sets? 2. Whenever you measure various quantities in an experiment there are uncertainties associated with these measurements. Refer to the uncertainty analysis instructions found online on Pilot to state which measurements of this experiment are considered systematic, and which are considered random. In your statement give reasons why they are systematic, and why they are random. 3. In part 2 you changed the attachment point at which the “Bicep” is attached to the forearm. Compare the measured “Bicep” force from part 1-e to the measured “Bicep” force from part 2. Include both values in your answer. Discuss the reasons why there is difference between the two values. 4. In part 1 you determined the vertical force on the “Elbow” for three different cases (no weight, 50-gram weight, and 100-gram weight). There is also a horizontal force on the “Elbow” that was not considered in this experiment. Describe how you would determine this horizontal force on the “Elbow” using the data that you already have.