Experiment on Force, Torque, and Equilibrium - Lab Report Example

Table of Contents Table of Contents A Report on the Experiment on Force, Torque and Equilibrium” 2 Purpose 2 Principle 2 Experiment Equipment and Material 2 Part one: Rotational Equilibrium of an object 3 Experiment procedure 3 Results, Discussion and Analysis 5 Conclusion 7 Part Two: Forces and torques in a Human Arm Model 8 Procedure 8 Results Discussions and Analysis 9 Conclusion 10 Work Cited 11 “A Report on the Experiment on Force, Torque and Equilibrium” Purpose The report will offer an insight on the relationship between force, torque and equilibrium. Diverse forces will apply at different positions and directions. The Equal Arm Balance positioned at a free point of angle and is to rotate about a fixed axis. The second requirement of equilibrium will be required to be confirmed that the net torque equals to zero. In addition, the experiment through the use of the Arm Model a particular force created by a biceps muscle will be determined due to its contraction, when the arm lifts an object whose weight is known. Principle The laws that are applied in this experiment are represented in the following formulas τ = rF sin θ Where τ represents torque, r= direction F= magnitude of force Torque refers to the rotational equivalent of force. It helps in measuring the efficiency of a force in an accelerating rotation. Torque is a three dimensional vector. Experiment Equipment and Material Computer Verneir Force Sensor Go! Link interface Equal Arm balance Rod with a table Clamp Electronic scale Protractor Mass set with a hanger Human Arm Model Part one: Rotational Equilibrium of an object Experiment procedure This experiment entails the measurement of the forces that are applied on an object. The object under consideration in this case is the equal arm balance. The torque created by individual forces will be computed and then we will show that the computed net torque is zero. 1. To compute torque due to individual force applied to the Equal Arm Balance at diverse positions in diverse directions, it is requisite to establish the direction of a vector, which is drawn from the axis. This brings us to the determination of the angle between r and F, which is given by the following equation θ = θ F - θ 1-3 2. Then create a table. Measure and record the height connecting the axis and the base of the Equal Arm Balance 3. Use equation two and three to compute the angle between r and the base of the Equal Arm Balance for individual positions 1, 2 and 3. 4. Link the force sensor to a Go-Link. Then plug it into the computer. Double click on Force and torque. This will facilitate opening of the LoggerPro Program. The program will be used to read and show the force sensor senor data. 5. Then hook a hanger with 150g mass from the left point indicated 3. Also a mass holder loop 6. Then hang a hook to the right side of the balance at point indicated 3. Then attach the force sensor to the other end of the cord. 7. Then hold the force sensor without pulling it. Press “Zero”. You will see the reading to be zero. 8. Pull the sensor downward. This will adjust the balance to the horizontal position. Then press “Collect”. During this time, the balance should rest at the equilibrium 9. Determine the mean from the graph of force versus time. At this point calculation of the net torque applied to the balance when at equilibrium can begin. 10. Hanging mass apply to the balance is equivalent to the weight of the mass. Consider the following equation W=Mg. the angle between force and the base = 90 degrees then we have the torque created to be as Mg·r3·sin θ = Mg·r3·sin (90°- θ 3), the torque is positive hence counterclockwise direction 11. The torque about the axis is given by F·r3·sin θ = F·r3·sin (90°- θ 3). Since the direction is clockwise, then the torque is negative. 12. 12. Compute the net torque given by θ net= θ W+ θ F. In addition compute the percentage error given by 13 and 14, repeat the computations using different positions 15. Find the average error 16. Conclude on the fulfillment of the 2nd condition of equilibrium on the experiment Results, Discussion and Analysis The results were computed from the data given Position h , m r, m Sin θ 1-3 =h/r 1-3 Θ 1-3 1r1 1.8 1.8 27.50 62.50 2r2 1.8 3.8 71.30 161.30 3r4 1.8 5.7 00 900 Sin θ 1-3 =h/r 1-3 =1.8-5.7= 3.9 Sine= 1.8/-3.9 =0.4615 Sine inverse =27.50 Sin θ 2-3 =h/r 2-3 3.88-5.7= -1.9 Sine =1.8/-1.9 =0.9474 Angle= 71.3o Sin θ 3-3 =h/r 3-3 , 5.7-5.7= 0 Angle = 00 Angle θ =30 o Calculations; Sin θ1 = h/r1 = 1.8/1.8= sine inverse of = 1 Angle =90 0 Sine θ2 = h/r2 1.8/3.8 = sine inverse of 0.4737 =28.30 Sine θ3 = h/ r3 =1.8/5.7 =0.3158 Sine inverse = 18.40 The angles between r and the horizontal arm balance are as follows R1 = 900 R2 28.30 R3 18.40 Mass holder loop position Mass in Kg Angle θ W, degrees Torque θ W, Nm Applied force loop position Force N Angle θ F, degrees Torque θ F, N·m Net torque θ net= θ W+ θF, N·m Percent error, % 3 left 0.15 90 8.11 3rigth 1.5 90 -8.11 0 0 3 left 0.15 90 8.11 3right 1.5 70 -6.7 1.41 17.38 3 left 0.15 90 8.11 2right 1.5 90 -8.11 0 0 3 left 0.15 90 8.11 2right 1.5 70 -6.7 1.41 17.38 1 left 0.15 90 8.11 3 right 1.5 90 -8.11 0 0 Torque = Mg·r3·sin (90°- θ 3). 150/1000(10N/kg).5.7sin (90-18.4) = 8.113Nm Average percent error = 17.38+17.38/2 = 17.38 Conclusion The fact that the net sum of vector torques adds up to zero confirms that at the point of equilibrium the object is not in directional movement that can lead to change in magnitude of distance but it is constrained to move in a rotational manner due to the fulfillment of the above two principles under investigation (net force equals zero and net vector sum of torque equals zero). In this regard, the principle of Isaac Newton’s force of gravity and centripetal acceleration are fundamentally essential to note. Hence the experiment affirms the fulfillment of the second condition of equilibrium. Part Two: Forces and torques in a Human Arm Model The science that is behind the functionality of the human arm is immensely attributed to the movement of the muscles which result to a motion supported by the biceps and the triceps of the arm. This section, will offer an opportunity for measuring the forces attributed to the biceps for holding a weight on the hand at 90 degrees. Computation of the torque due to this force comparative to the weight model that the arm holds will be done. Procedure 1. The force sensor should be fastened on the vertical rod just right on top of the arm model. Then one end of the cord attached to the insertion point of a standard biceps. Then direct the cord and attach it to the other end of the force sensor. At this point adjust the length and the sensor location to a 90 degree angle. This is between the elbow and the shoulder which is our vertical. 2. loosen the cord by lifting the arm in a s low pace, then press Zero to enable you bring the force reading to the zero point. This is an indication of “no load connected”. 3. Allow the arm to swing and then record the force of the biceps as Fb1. This force is requisite to hold the arm in a horizontal manner without adding any load 4. Now a load of 100-g can be added to the arm model’s hand. Record this force as Fb2. This is the force requisite to hold the load weighing 100g at the horizontal position. 5. Then measure the distance from the point where a force W= Mg is applied and the middle of the load 6. Assess the distance between the point of rotation and the center of the load. Account this as rw. 7. Compute the torque that is due to the weight of the mass. This is given by the equation θ w = Mg· rw 8. Make a computation of the extra torque attributed to the biceps force, (tb = (Fb2 –Fb1). rb). In this case the direction is counterclockwise which is positive. 9. The experiment should indicate that the two torques balance each other. This should happen because the equilibrium represented is rotational 10. Make conclusions on the magnitude of the biceps force comparative to the weight of the hand. Results Discussions and Analysis Let the biceps force be Fb1 when there is no load When a mass of 100 grams is added, the force is measured to be Fb2 Let the distance between the standard insertion point and the rotation point, be rb i.e. the distance from the point of rotation to the point where the biceps force is applied to the arm Let the distance between the rotation point and the center of the load (the point where the force w = Mg is applied and be as rw. The calculation of torques is thus; Torque due to biceps force is tb = (Fb2 – Fb1) ·rb. Fb2-Fb1=F Therefore torque =F*rb, since rb equals 5.7cm then the actual torque is 5.7*F=5.7F sin90 This equals 5.7F in the positive direction Torque due to the weight = tw = Mg·rw. But mg=mass *gravity = 100/1000(10N/Kg) =force F Since distance due to mass = rw=5.7cm, ten torque=5.7F but in the negative direction. The torqueses are equal. Conclusion Since the two torqueses are equal and their sum equals zero as well as their net forces satisfying the two conditions of rotational equilibrium. Thus, the expectations of the two experiments are met. Work Cited Wilson, Jerry D, and Cecilia A. H. Hall. Physics Laboratory Experiments. Boston, MA: Brooks/Cole, Cengage Learning, 2009. Print. Page 193 Read More