ソース: ミッチェル ・ ウィン博士は Ketron、Asantha Cooray、PhD、物理教室 & 天文学、物理的な科学の学校、カリフォルニア大学、アーバイン、カリフォルニア州
平衡は、日常生活の中で非常に重要な力学の特殊なケースです。それは、正味の力とオブジェクトまたはシステムの純トルクが両方とも 0 の場合に発生します。これは、長さと角度の加速度がゼロであることを意味します。したがって、オブジェクトは、安静時や一定速度でその重心を移動します。ただし、システム内のオブジェクトに対する行動している力がないわけではないです。実際には、任意のオブジェクトに力が作用しない地球上非常にほとんどのシナリオがありません。人は、橋を渡って歩く、その質量に比例して橋の上の下向きの力を出すし、橋同等を発揮し、人に上向きの力の反対。いくつかのケースで橋の人の下向きの力への応答で屈曲性がありますや極端な場合、力が十分に大きいとき橋真剣に変形になることがあります可能性がありますも破壊します。この平衡のオブジェクトの屈曲の研究弾力性と建物及び構築物私達が毎日を使用するエンジニアを設計しているとき非常に重要になります。
The representative results for the experiment can be found in Table 1. The force exerted on the two springs by the hanging mass are denoted by their locations: left and right, denoted by subscripts L and R. Since there are two unknowns in this experiment, FLand FR, two equations are required to solve for them. Thus, Equations 1 and 2 are used to solve for the two forces. The torques are used to obtain a relationship between FLand FR .
Since the force exerted by the weight is downward, the angle θ in Equation 3 is 90°,and the torque is just r · F. The torques τLand τR are also in opposite directions, where counterclockwise is defined as the positive direction. Using Equation 2
–τL + τR = 0 = –rL FL + rR FR. (Equation 5)
Equivalently,
FL = FR rR/rL. (Equation 6)
Using Equation 1
FL + FR = m g, (Equation 7)
where m is the mass of the weight and g is the gravitational constant of 9.8 m/s2. In other words, the downward force of the weight equals the sum of the forces holding up the weight and meter stick system, which is just the two springs on the left and right, which are suspending the system. With these two equations (6 and 7), the unknowns FL and FR can be calculated. These are shown in Table 1. These values are compared with the forces exerted on the springs in the last two columns of the table. Slight discrepancies are expected from measurement errors. In addition, it has been assumed that the mass of the meter stick is zero, which is incorrect, strictly speaking, but nevertheless a good approximation. This lab uses spring scales, which show how many Newtons are being applied to the spring when stretched, so it is not necessary to know the spring constant, k.
Table 1. Theoretical and experimental results.
| Mass (g) | rL (cm) | rR (cm) | FL (N) | FR (N) | FL,spring (N) | FR,spring (N) | % diff (left) | % diff (right) |
| 100 | 50 | 50 | 0.5 | 0.5 | 0.45 | 0.45 | 9.9 | 9.9 |
| 100 | 30 | 70 | 0.68 | 0.29 | 0.65 | 0.3 | 4.4 | 3.4 |
| 100 | 10 | 90 | 0.9 | 0.1 | 0.85 | 0.1 | 5.5 | 0 |
| 200 | 50 | 50 | 0.98 | 0.98 | 1 | 1 | 0 | 0 |
| 200 | 30 | 70 | 1.38 | 0.59 | 1.35 | 0.55 | 2.1 | 7.2 |
| 200 | 10 | 90 | 1.8 | 0.2 | 1.85 | 0.2 | 2.7 | 0 |
All bridges are under some amount of stress, from both their own weight and the weight of the loads moving across. Suspension bridges, like the Golden Gate, are a complex system of objects under very heavy forces and in equilibrium. The cables that hold the bridge up are elastic, and their elasticity was considered when the structural engineers designed the bridge. Similarly, skyscrapers have a complex system of steel beams under tremendous forces, which altogether compose a rigid system in static equilibrium. Elasticity plays a role in the materials used to construct buildings, as they need to be able to withstand a certain amount of flexing, especially in areas where earthquakes are prevalent. The cranes used to construct these structures are also in equilibrium, with a complex system of cables and pulleys to lift and lower the construction materials.
In this study, the equilibrium of a system composed of multiple components under various forces was observed. The effects of the elastic components were also observed using spring scales of known spring constants. The forces exerted upon the springs were computed using the two conditions necessary for equilibrium: the sum of the forces and the sum of the torques are zero.
Equilibrium is a special case in classical mechanics but is ubiquitous in everyday life, while free-body diagrams help decipher the underlying forces present.
A system is in translational equilibrium if the forces acting on it are balanced, that is, the net force is zero. Equilibrium can also be established in a rotational system if the net torque, t, is zero.
In addition to these static equilibrium cases where the systems are at rest, dynamic equilibrium implies that a system is moving but experiencing no linear acceleration a or angular acceleration, a.
Now, even if a system is in equilibrium, a multitude of individual forces or torques can be acting on it, and free-body diagrams — composed of simple shapes and arrows — are often implemented in order to conceptualize these forces and/or torques acting on a system.
The goal of this experiment is to understand the equilibrium of a system composed of multiple components under the influence of various forces.
Before analyzing this complex system, let’s revisit the concepts of equilibrium and free-body diagrams. As mentioned earlier, equilibrium occurs in a translational system, such as a loaded spring, when the restoring force balances the gravitational weight. In a rotational system, example when weights are attached to a freely rotating beam, equilibrium is established when the torques balance one another. Note that, with respect to the axis of rotation, torque is positive for counter-clockwise rotation and negative for clockwise rotation.
In these cases, the net forces or torques are equal zero and therefore no linear or angular acceleration exists. Per Newton’s First Law, since these systems are in static equilibrium they must remain at rest.
Despite the absence of a net force or torque, multiple forces are acting on the objects within these systems. Free-body diagrams, or force diagrams, are often drawn in order to understand the forces and torques acting on systems in equilibrium.
Each contributing force or torque is represented by an arrow whose size and direction fully describes the vector in question. Through vector addition, the translational system is shown to be in equilibrium. Similarly, by accounting for the torque direction with respect to the axis, the rotational system is also in equilibrium.
Now, imagine combining these systems such that a weight is attached to the center of the beam while the beam itself is suspended at its ends by two springs. The system is complex but can be understood by using two separate free-body diagrams. The translational system includes the weight and the left and right spring restoring forces, denoted as FL and FR, respectively.
Since the system is in equilibrium, the sum of magnitudes of FL and FR should be equal to the magnitude of the weight. This equation describes transitional equilibrium.
In the rotational system, instead of forces we have torques. Recall that torque is defined as the perpendicular force times the distance r that the force is applied from the axis of rotation. Since the weight is positioned at the axis of rotation, it exerts no torque on the beam. Whereas for the springs in this case, the perpendicular forces are the restoring forces and r is the respective distance from the weight.
Now again, since the system is in equilibrium, the magnitudes of these torques should be equal, and this equation illustrates rotational equilibrium.
Moving the weight away from the center causes the beam to tilt. For the translational system, the sum of the restoring forces is still equal and opposite to that of the weight. Therefore, the equation for translational equilibrium — dealing with the magnitude of these forces — stays the same.
For the rotational system, the tilt by an angle θ changes the forces in the spring torques to the cosine component of the respective restoring forces. The lengths of the rotational arms also change. However, the weight is still at the axis of rotation and therefore exerts no torque on the beam.
Since this system is also in equilibrium, the magnitudes of the torques applied by the springs should be the same. Cancelling out the cosine θ, results in the same rotational equilibrium formula.
Now that you understand the principles of equilibrium, let’s apply these concepts to a system that experiences both forces and torques. This experiment consists of a meter stick, two spring scales, two stands, and two weights of different mass capable of being suspended from the meter stick.
To begin, place the two stands one meter apart on the table making sure they are secure. Suspend a spring scale off of each stand, and attach each end of a meter stick to the bottom of a spring scale.
Next, attach the least massive weight to the meter stick midway between the spring scales. With the system under translational and rotational equilibrium, calculate the individual forces acting on the meter stick and record them.
Read the values on each of the spring scales and record these restoring forces exerted by the springs.
Now shift the weight 0.2 m to the left making the left rotation arm 0.3 m and the right rotation arm 0.7 m. Repeat the calculation of the individual forces and the spring scale measurements.
Lastly, shift the weight to the left an additional 0.2 m and perform the force calculations and spring scale measurements. Repeat this equilibrium experiment for the more massive weight.
The individual forces acting on the meter stick consist of the gravitational force on the attached weight and the restoring forces of the springs. When looking at the free-body diagrams of the system under translational and rotational equilibrium, two equations can be used to determine the two unknown restoring forces.
The rotation arms are identical when the weight is midway between the springs. Therefore, each of the restoring forces should equal half of the weight. For the experiments when the weight is moved from the center, the restoring forces are dictated by the ratio of their respective rotation arms.
These calculated values can be compared with the restoring forces determined from the spring scale measurements. The differences between the values are within the measurement errors of the experiment. Therefore, by invoking equilibrium conditions, the restoring forces can be determined with knowledge of the mass of the weight and the length of the rotation arms.
The basic principles of equilibrium can be invaluable when engineers are designing structures that we use every day.
A bridge is always in static equilibrium while constantly experiencing large forces and torques from both its own weight and the loads moving across it. Therefore, construction of a suspension bridge, like the Golden Gate in San Francisco, requires significant structural engineering efforts to ensure that equilibrium is maintained even during the times of heavy traffic
Similarly, skyscrapers have a complex system of steel beams under tremendous forces, which altogether compose a rigid system in static equilibrium. Therefore, an understanding of the concepts behind equilibrium helps an architect decide the construction parameters, so that these structures can withstand a certain amount of torque, especially in the earthquake prone zones.
You’ve just watched JoVE’s introduction to Equilibrium. You should now understand the principles of equilibrium and how free-body diagrams can be used to determine the forces and torques contributing to a system in equilibrium. Thanks for watching!
