Forces and Torques in Equilibrium-3

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Apr 3, 2024

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Report for Experiment #8 Forces and Torques in Equilibrium Abstract This experiment primarily aimed to look at forces and torques in equilibrium. In both Investigations 1 and 2, this was accomplished by identifying an object’s center of gravity, measuring the torque a force produces around an axis, demonstrating that an object is at rest when all forces and torques add to zero and then using these concepts to describe an arm lifting a weight. It has been demonstrated through the construction of an experimental apparatus that when the applied force increases, the force’s distance from the center of gravity must also decrease in order to keep a constant torque. The experimentally obtained values, however, did not agree with one another, which could have been caused by a flaw in the apparatus construction or by subpar tools.

Introduction This lab’s investigation of torques and forces on an item in equilibrium was a key component. When an object is in equilibrium, all of the forces operating on it are equal to zero and it is at rest. The item will start to spin around an axis if the sum of the forces varies in strength, direction, or lever arm (the separation between the pivot and line of action). When this happens, the torque’s strength can be expressed as: τ = ࠵? ⊥ ࠵? F stands for the force applied to the item, r represents the lever arm, and shows that r is ⊥ parallel to the line of action. The force direction line going through the location where the force is applied is known as the line of action. Torque has an associated direction because it is a vector. In the following experiment, the torque were perpendicular to the lever arm along with force that produced throughout the entire Investigation. Additionally, a torque is typically thought of as positive if it rotates counterclockwise rather than negatively if it rotates clockwise. A simple technique for concluding the direction of the torque is the right hand ruler, which consist of extending your right hand in front of you with your finger point away from your body. Following that curl your fingers in the direction the object's rotation while pointing your thumb in the direction of torque. The requirement that the total force exerted on an extended object be zero is critical to remember ࠵? ࠵? = 0 Total of its torques applied to it must also equal zero: τ ࠵? = 0 When both of these conditions are met, the object is said to be in a static condition. In Investigation1, a weighted meter stick’s center of gravity was determined by balancing it on a finger. The finger had to be put closer to where the weight was since the meter stick’s mass was unevenly distributed as a result of the extra weight. As a result, the net torque values were acquired. The position of the finger was then used to determine the gravity’s center. Investigation 1 Finding the center of gravity on a weighted meter stick was the objective of this experiment. The stick’s center of gravity , which was calculated by balancing it on a finger, ࠵? ࠵?࠵? was found to be 0.426 meters, with a roughly accurate error of 0.0005 meter —equivalent to half the finger’s width. After that, it was discovered that the meter stick’s mass , was ࠵? 1 It was next attempted to balance the meter stick on a triangle pivot, but 0. 2804 ± 0. 00005࠵?࠵?. this proved to be quite challenging because it had a considerably smaller surface area than the finger employed initially. Table 1 below shows the information that was discovered. Table 1: Measurements of mass and the gravitational center, alongside the corresponding errors ࠵? 1 (࠵?࠵?) δ࠵? 1 (࠵?࠵?) ࠵? ࠵?࠵? (࠵?) δ࠵? ࠵?࠵? (࠵?) 0. 2804 0. 00005 0. 426 0. 0005

The outcomes of this experiment demonstrate that the center of gravity is not the actual physical center of an object but rather a unique location where an object may balance because its net torque is zero. If the center of gravity is known, any further computations can be considered as being aided at this point by the weight of the entire meter stick. Investigation 2 The Introduction and the lab manual provided descriptions of the apparatus utilized [1]. First, a frame was constructed using two-rod stands, a long rod, and a hook collar. The triangular pivot of the short rod, which had one sharp edge projecting out, was clamped to the left of the vertical rod. The hole on end closest to the zero of the meter stick was then used to install it on the pivot. The long top rod had a pulley attached to it. A string was looped around the meter stick, passed through the pulley, and then tied to a mass hanger. The pivot was not exactly at zero on the meter stick, therefore there was a very small inconsequential gap which was roughly 0.308m between the string’s location and it. By dividing the pivot’s width by 2, it was possible to calculate the inaccuracy for the distance . The pulley’s position was altered δ࠵? ࠵? to allow the string to rise vertically. Along with the magnitude of the “biceps” force, or the force applied by the string to ࠵? ࠵? hold the sick horizontal, the distance and (equivalent to half the width of pivot) were also ࠵? ࠵? δ࠵? ࠵? noted. It was calculated by dividing the sum of the hanging weights’ total masses by the gravitational force. Similarly the error of hanging weights’ was calculated by dividing the δ࠵? ࠵? small unit of mass by half. was determined by multiplying by gravity. In light of the fact δ࠵? ࠵? δ࠵? ࠵? that the force exerted on the meter stick was perpendicular, Eq. 1 was then used to compute the inaccuracy about the pivot ( . The arm is pulled in a positive direction and arms motion is in a τ ࠵? ) counterclockwise direction , hence the torque has a positive value. The relative error of the “bicep” was computed using; . The computed value was used to find relative error of the δ࠵? ࠵? ࠵? ࠵? “biceps” torque: . δτ ࠵? τ ࠵? Table 2: Measurements from part one of Investigation 2. ࠵? ࠵? (m) δ࠵? ࠵? (m) ࠵? ࠵? (kg) δ࠵? ࠵? (kg) ࠵? ࠵?࠵? (N) δ࠵? ࠵?࠵? (N) δ࠵? ࠵?࠵? /࠵? ࠵?࠵? τ ࠵? (Nm) δτ ࠵? (Nm) δτ ࠵? /τ ࠵? 0. 308 0. 005 0. 4 0. 0005 3. 92 0. 0049 0. 000125 1. 207 0. 001425 0. 00118 ࠵? 1 (kg) δ࠵? 1 (kg) δ࠵? 1 /࠵? 1 (%) τ ࠵?1 (Nm) δτ ࠵?1 (Nm) δτ ࠵?1 /τ ࠵?1 ࠵? ࠵? (N) 2. 74792 0. 00049 0. 0178 1. 170614 0. 02091917 0. 1187 − 1. 17208

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To replicate the hang-carrying weight for the second phase of this research, a 0.005kg mass hanger was placed on the other of the meter stick, by increasing the mass of the ࠵? 2 hanger, the biceps force was modified to bring the meter stick back to a horizontal by measuring the position from the fulcrum . The new total of the suspended mass was determined and ࠵? ࠵?2 expressed as the following by multiplying the mass by gravity: . In order to calculate a new ࠵? ࠵?࠵? torque value , this was then applied to Eq .1. The ratio of the increase in biceps force to τ ࠵? ∆࠵? ࠵? the weight that the hand is lifting: ࠵? 2 ∆࠵? ࠵? ࠵? 2 = (࠵? ࠵?࠵? −࠵? ࠵?࠵? ) ࠵? 2 By multiplying by , the amount of the second weight force’s torque was calculated. ࠵? ࠵?2 ࠵? 2 τ ࠵?2 This was contrasted with the rise in biceps torque magnitude was found to be ∆τ ࠵? = ∆࠵? ࠵? × ࠵? ࠵? marginally larger. Tables 1 and 2 showcase the provided information that was discovered throughout the Investigations. Table 3: Measurements from part two of Investigation 2. ࠵? 2 (kg) ࠵? ࠵?2 (m) ࠵? ࠵?࠵? (N) δ࠵? ࠵?࠵? (N) ∆࠵? ࠵?࠵? (N) τ ࠵?2 (Nm) τ ࠵? (Nm) ∆τ ࠵? 0. 49 0. 98 5. 54386 0. 00049 1. 62386 0. 4802 1. 6508 0. 500149 Experimental results revealed that the first was Nm and the first τ ࠵? 1. 207 ± 0. 001425 τ ࠵? was Nm. It is evident that these numbers are out of range when 1. 170614 ± 0. 02091917 compared to their experimental errors. SInce they must add up to zero in order for the meter stick to be at rest, this goes against the expectation that they would be equal within their ranges. This mistake might have happened as a result of improper construction of the experimental structure or a slightly bent pivot that prevented the meter stick from ever being perfectly horizontal. The experimentally determined value of in the investigation’s second phase was τ ࠵?2 0.4802 Nm, which was extremely near to the value of , which was discovered to be ∆τ ࠵? 0.500149. Since the torques should balance out to allow the item to be at rest, this is to be expected. The tiny discrepancy in the data might have been caused by the same issue that was previously addressed. Conclusion The objectives of this experiment were to locate an object’s center of gravity, demonstrate that an object’s forces and torques add to zero when it is at rest, calculate the torque applied by a force about an axis, and apply these concepts to an arm lifting a weight. A meter stick was balanced on a finger to study these concepts, and a device to replicate a “bicep” maintaining an arm straight with and without additional weight in the “hand” was made. The weights required to keep the meter stick horizontal were utilized to calculate the torques involved in the structure as well as forces pressing on it.

When the weighted meter stick in Investigation 1 was balanced on a ginger at a distance of 0.426 m, all of the stresses and torques acting on the stick were balanced and added up to zero. This was the location of the center of mass or gravity Without any additional weight, Investigation 2’s results showed that the biceps’ torque, measured as , and the meter stick’s weight, measured as 1. 207 ± 0. 001425 ࠵?࠵? , respectively. These two values do not correspond to one another, 1. 170614 ± 0. 02091917 ࠵?࠵? which may be the result of an error in the experimental apparatus’ construction or a slightly bent meter stick pivot. The additional torque caused by the weight of the meter stick, , was τ ࠵?2 calculated experimentally to be 0.4802 Nm. This value is extremely similar to the increment in the magnitude of the bicep’s torque , which was determined to be ∆τ ࠵? 0. 500149 Question Explain why balancing the forces acting on a body is not enough to establish equilibrium. Give an example to justify your answer. Because balancing forces only affect an object’s movement in one of four directions -up, down, left, or right-rather than its rotation, a body cannot be in equilibrium with balancing force alone. A dancer spinning in place would serve as an illustration of this. She is not in equilibrium because she is spinning, but because she is not moving, the forces pulling on her are equal and balanced. Explain why balancing the torques acting on a body is not enough to establish equilibrium. Give an example to justify your answer. Because torques only affect an object’s rotation and do not affect its up, down, left, or right movement, balancing their effects on a body is insufficient to achieve equilibrium. A moving elevator would be one illustration of this. Although the torques operating on it are balanced, because it is still moving up and down, it is not in equilibrium. Why are door handles usually located as far as possible from the door’s hinge? Door handles are usually located far from the door’s hinge since that is the pivot point of the door. If they were located closer, the r value in the torque equation would approach zero, making the torque magnitude smaller, and the door harder to open. 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? It is easy to lay a pen horizontally on a table because the mass is equally distributed throughout the whole pen. When you attempt to balance it vertically on the tip, all of the pen’s mass must be distributed through a small point, so it is very hard to keep it balanced. Why do rowers typically have the same number of paddles on each side of the boat? In order to ensure that the forces and torque acting on both sides of the boat are equal, rowers normally have an equal number of paddles on each side of their boat. The boat will only turn in one direction if one side has paddles, but if both side have the same number, the torque values are identical and the force can only move the boat forward.

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