PHYS1146 Lab 4

Report for Experiment #8 Forces and Torques in Equilibrium 10/20/2023 Abstract The forces and torques acting on a meter stick was investigated with and without added weight. This was done using a set-up that simulated the lower part of the arm. In the first investigation the center of gravity of a weighted meter stick was determined to be 0.405±0.0005 m . This meter stick was then used in the second investigation and the force reqired in the simulated bicep to maintain equilibrium was determined and the resulant torque around the pivot point was calculated. Without the weight the bicep force was 10.310±0.0005 N and the bicep torque was 1.206±0.005 N•m . With weight the bicep force was 14.230±0.0005 N and the bicep torque was 1.665±0.007 N•m. The increase in bicep torque was very similar to the increase in torque from adding the weight. A systematic error affecting the distance at which the bicep string was attached was identified. Introduction Torque is a type of force that causes rotation of an object and is defined as the force multiplied by the distance from the pivot point with the units being N•m . The rotation is about a point which is called the pivot point. This is the center point of a system that rotates. According to Newton’s first law when an object is at equilibrium all the forces acting upon the object must sum to zero. However, an object can still

rotate about a pivot point when the sum of forces is zero, if there is an unequal torque applied. For an object to be completely at rest both the sum of the forces and the sum of the torques must be zero. The goal of this experiment was to find the center of gravity of a meter stick and simulate the forces acting on the lower part of a human arm holding a weight in the hand. The elbow acts as a pivot point and bicep muscle acts as an upward force balancing the force exerted on the weight by gravity. By doing this the torque applied by the weight could be calculated and the corresponding torque that the biceps needed to apply to maintain the forearm at complete rest could also be calculated. The expectation was that the force provided by the bicep was expected to be higher than the gravitational force exerted on the weight because the bicep was acting at a much shorter distance from the pivot point than the weight. Investigation 1 The experimental setup consisted of a weighted meter stick, digital scale, finger, and triangular pivot. The weight on the meter stick caused the mass to be unevenly distributed, this caused the center of gravity to not be centered. The center of gravity had to be found by figuring out where the meter stick would balance and not rotate. When the meter stick did not rotate the torque was zero. The meter stick was first balanced on a finger. The position of the finger was adjusted until the meter stick did not rotate. The position of the center of gravity was found by seeing at what point on the meter stick the finger was when the meter stick did not rotate, the value used was the mark at the finger’s estimated halfway point. The uncertainty in the center of gravity was found by halving the smallest value of measurement on the meter stick. The mass of the meter stick was then found by using a digital scale. The error in mass was found by halving the smallest value of measurement on the digital scale. The meter stick was then balanced on the triangular pivot and the point at which the ruler balanced was recorded. Table 1: Mass, center of gravity, and weight of meter stick. The mass of the meter stick was found to be 0.310±0.00005 kg . The center of gravity (r1) was found by finding where the meter stick did not rotate. The position where the meter stick did not rotate on both the finger and triangular pivot was at the 0.405±0.0005 m mark. The weight of the meter stick was found by multiplying the measured mass by gravity, W1= m1 × 9.8 , and was 3.038±0.00005 N . The error in weight was the same as the error in mass because W1=m1 × g and there is no error in the value of gravity so the propagated error function for multiplied values would be the same as error in mass. The center of gravity was not the center of the meter stick due to the presence of a mass towards one end of the ruler. The balance point was found with both a finger and triangular pivot. Both values were the same at 0.405±0.0005 m . The triangular pivot point was expected to give a more accurate reading because the pivot point was narrower than the finger. The results were consistent and expected. m1 (mass of meter stick) (kg) 0.310 ϭ m1 (kg) 0.00005 r1 with finger (m) 0.405 r1 with pivot (m) 0.405 ϭ r (m) 0.0005 W1 ( N ) 3.038 ϭ W 1 ( N ) 0.00005

(1) The relative error in the bicep torque was found by dividing the error in torque by the calculated value of the bicep torque. The weight of the meter stick was calculated in investigation one and acted as a downward force on the meter stick. Torque exerted by the weight was found multiplying the mass of the meter stick by the distance of its center of gravity from the pivot point. The value of torque was found to be –1.206± 0.002 N•m . The error in torque was calculated by using (1) with the mass and error in mass of the meter stick and the center of gravity and error in center of gravity. These values were found in investigation one. The value of the torque exerted by the weight of the meter stick was negative because it was a downward force, while the force of the biceps was an upward force and positive. The magnitude of torque due to the bicep was equal to the magnitude of the torque due to the weight of the meter stick. This was because the meter stick was in equilibrium, therefore the sum of the forces equals zero and this was expected. These values are equal and therefore within the limits of experimental error. The precent errors of weight force and torque of the meter stick were calculated by dividing their error value by their calculated value and then multiplied by one hundred. The percent error of weight force was 100, which was equal to 0.002%. The precent error of the torque of the meter stick was 100, which was equal to 0.127%. The force on the meter stick due to the pivot was found by subtracting the bicep force by the force due to the meter stick's center of gravity. Because the meter stick was at equilibrium at that moment the sum of forces had to be zero which meant the force due to the pivot was –7.272±0.0005 N . However, because the force due to the pivot was directly acting on the pivot, the distance from the pivot was zero and had no influence on the torque. Then a second mass hanger was hung at the very end of the meter stick and was measured to be 50 g . The distance this mass hanger was from the pivot was found to be 99.2 cm and then additional weights were added to the first mass hanger to bring the meter stick back to a horizontal position. The force from the added weight was 0.490±0.00005 N . This was found by multiplying the mass of the mass hanger by gravity. The error was equal to the error in mass because there is no error in the value of gravity so the propagated error function for multiplied values would be the same as error in mass. The new bicep force required was 14.230±0.0005 N . The force was calculated by multiplying the mass of the weights (1.452±0.00005 kg ) by gravity. The error in mass was half of the smallest measurement value on the meter stick. The new torque was 1.665±0.007 N•m . The torque was calculated by multiplying the new bicep force by the distance the string was from the pivot. The error in torque was calculated by using (1). The change in bicep force was 3.920±0.001 N , this was found by subtracting the new bicep force by the old bicep force. The error of the change in bicep force was calculated using: (2) The ratio of increase was found to be 8.000. This was calculated by dividing the change in bicep force by the force from weight two. The magnitude of the torque from weight two was calculated by multiplying the force from weight two by the distance from the pivot point, 0.490 × 0.992 m = 0.486±0.0005 N•m . The error in the torque from weight two was calculated by using (1). The biceps torque increased in magnitude by 0.459 N•m. This was calculated by subtracting the old bicep torque from the new bicep torque. After weight two was added the bicep required more torque in order to maintain equilibrium. The torque from weight two and the increase in torque was supposed to be identical so that all torques acting on the meter δτ = ( δ F F ) 2 + ( δ r r ) 2 ( τ ) δ W 1 W 1 × δτ w 1 τ w 1 × δ ∆ F s = 2 × ( δ F s ) 2

stick are equally balanced. The two torques measured were very similar but were not within the range of error. Table 2: Forces, distances, and torques from investigation two. Table 3: Table from investigation one with % errors with the CG changed to distance from pivot point. rs (m) 0.117 ϭ rs (m) 0.0005 mass of weights (kg) 1.052 old Fs( N ) 10.310 old ϭ Fs( N ) 0.0005 τ s ( N m) 1.206 ϭτ s ( N m) 0.005 relative error in the biceps torque 0.004 relative error in the biceps force 0.000 τ w1 ( N m) -1.206 ϭτ w1( N m) 0.002 Fp ( N ) -7.272 new r w2 (m) 0.992 N ew mass of the weights (kg) 1.452 N ew Fs ( N ) 14.230 new τ s ( N m) 1.665 ∆ Fs ( N ) 3.920 ϭ∆ Fs ( N ) 0.001 W2 ( N ) 0.490 ϭ W2 ( N ) 0.00005 Ratio of ∆ Fs to W2 8.000 τ w2 ( N m) -0.486 ϭτ w2 ( N m) 0.0005 ϭτ s ( N m) 0.007 ϭ∆τ s ( N m) 0.002 m1 (mass of meter stick) (kg) 0.310 ϭ m1 (kg) 0.00005

2. Explain why balancing the torques acting on a body is not enough to establish equilibrium. Give an example to justify your answer. a. When torques are zero, forces can still act upon an object and cause linear movement. For example, a rod with two equal weights on each end will accelerate when dropped but will not rotate because the torque forces are balanced. 3. Why are door handles usually located as far as possible from the door’s hinge? a. This is because doors open on a pivot at one end and use torque to open, therefore the further from the pivot the less force is required to generate the necessary torque to open the door. 4. While it is easy to lay a pen horizontally on a table, it can be exceptionally difficult to balance it vertically on its narrow end. Why? a. In order for the pen to balance on the narrow end the center of gravity must be perfectly aligned over the point of contact with the table. If there is not perfect alignment torque will be applied to the pen and it will fall, getting perfect alignment is very difficult. When the pen is on the table horizontally the contact point will include the center of gravity and equal force will be applied throughout the object and no torque will be applied. 5. Why do rowers typically have the same number of paddles on each side of the boat a. This is in order to prevent a torque from developing, which would turn the boat off a straight path. Honors Questions 1. Can a solid object have its center of gravity located outside of its solid boundary? a. Yes. For example, a metal ring is a solid object, however the center of gravity is close to the center of the ring where there is no material. 2. A construction worker can move a very heavy concrete block by using a metal crowbar. Which physics principle allows him to do this? Explain. a. The principle of torque, the force applied at the long end of the metal crowbar is less than that needed to raise the concrete block at the short end of the crowbar as torque is force multiplied by distance from pivot point. 3. Cavemen moved heavy stones, which they could not lift, and they had no crowbars. How did they do this? (In a similar way ants move objects that they cannot lift.) a. Ants move objects by pulling them along the ground in a team. Cavemen would act similarly to move heavy objects. By working in a team their forces would add together to

move the object. In addition, some scholars believe cavemen used wooden rollers to roll the heavy object along the ground. 4. Assume that the mass of the pulley is 12 g in your second Investigation. Calculate the force that acts on the hook that holds it (refer to Fig. 8.4). a. ; = 14.463 N b. ; = 29.044 N Acknowledgments References [1] Hyde, Batishchev, and Altunkaynak, Introductory Physics Laboratory, pp 58-64, Macmillan Higher Education, 2022. [2] IPL Lab Report Guide, https://web.northeastern.edu/ipl/wp-content/uploads/2017/09/IPL-Lab-Report- Guide.pdf [3] Giancoli, Physics Principles with Applications, Pearson, 2021. F s × r s = ( w 1 × r cg ) + ( w 2 × r w 2 ) F s = (3.038 × 0.397) + (0.49 × 0.992) 0.117 F = 2 F s + ( m p × g ) F = 2 ( 14.463 ) + (0.012 × 9.8)