Solving Static Equilibrium Problems Examples: The diagram shows a beam hinged at P. The force of gravity on the beam is 425N. Transcribed Image Text The conditions for static equilibrium are... (select all that apply) The sum of all torques is zero. It is in equilibrium both translationally and rotationally. These two conditions for equilibrium result in a system that has no tendency to accelerate linearly, or angularly. With this choice we only need to write Equation \ref{12.7} and Equation \ref{12.9} because all the y-components are identically zero. If you tilt a box so that one edge remains in contact with the table beneath it, then one edge of the base of support becomes a pivot. The free-body diagram for a body is a useful tool that allows us to count correctly all contributions from all external forces and torques acting on the body. A person of mass Mere Stands three-quarters of the way up from the bottom of the ladder. [/latex], [latex]\sum _{k}{\mathbf{\overset{\to }{\tau }}}_{k}=I\mathbf{\overset{\to }{\alpha }}. When using Equation \ref{12.10}, we often compute the magnitude of torque and assign its sense as either positive (+) or negative (−), depending on the direction of rotation caused by this torque alone. Springs If we set the acceleration to zero in Figure, we obtain the following equation: The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium: The first equilibrium condition, Figure, is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws. The first equilibrium condition, (Figure), is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws. where [latex]{r}_{k}[/latex] is the length of the lever arm of the force and [latex]{F}_{k}[/latex] is the magnitude of the force (as you saw in Fixed-Axis Rotation). Suppose now that the market is disturbed by a rightward shift of the demand curve from D 0 D 0 to D 1 D 1. However, to be complete the net torque must also be zero – called rotational equilibrium. In the final step in this chain of reasoning, we used the fact that in equilibrium in the old frame of reference, S, the first term vanishes because of Figure and the second term vanishes because of Figure. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). This means that a body in equilibrium can be moving, but if so, its linear and angular velocities must be constant. The conditions for equilibrium are basic to the design of any load-bearing structure such as a bridge or a building since such structures must be able to maintain equilibrium under load. In free-body diagrams, the weight vector is attached to the center of gravity of the body. By setting to zero the right-hand side of Figure, we obtain the second equilibrium condition: The second equilibrium condition for the static equilibrium of a rigid body expresses rotational equilibrium: The second equilibrium condition, Figure, is the equilibrium condition for torques that we encountered when we studied rotational dynamics. The first equilibrium condition, Equation 12.2.2, is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws. Again, this vector equation is equivalent to three scalar equations for the vector components of the net torque: The second equilibrium condition means that in equilibrium, there is no net external torque to cause rotation about any axis. Only in situations where a body has a large spatial extension so that the gravitational field is nonuniform throughout its volume, are the center of gravity and the center of mass located at different points. View this demonstration to experiment with stable and unstable positions of a box. The smaller boy on the right has a mass of 40.0 kg. In Figure, the z-component of torque [latex]{\mathbf{\overset{\to }{\tau }}}_{k}[/latex] from the force [latex]{\mathbf{\overset{\to }{F}}}_{k}[/latex] is. Legal. Give an example. The standard procedure is to adopt a frame of reference where the z-axis is the axis of rotation. For different choices of the pivot point we have different sets of equilibrium conditions to solve. Static equilibrium is the opposite. A passenger car with a 2.5-m wheelbase has 52% of its weight on the front wheels on level ground, as illustrated in Figure 12.4. Hence, the netF = 0 is a necessary—but not sufficient—condition for … For example, in the case of a tipping truck (Figure \(\PageIndex{2}\)), the pivot is located on the line where the tires make contact with the road’s surface. When we use the first equation to eliminate [latex]{T}_{2}[/latex] from the second equation, we obtain the relation between the mass [latex]m[/latex] on the pan and the tension [latex]{T}_{1}[/latex] in the shorter string: The string breaks when the tension reaches the critical value of [latex]{T}_{1}=2.80\,\text{N}. When using Figure, we often compute the magnitude of torque and assign its sense as either positive [latex](+)[/latex] or negative [latex](-),[/latex] depending on the direction of rotation caused by this torque alone. If the CM is located high above the road’s surface, the gravitational torque may be large enough to turn the truck over. 2.2 Coordinate Systems and Components of a Vector, 3.1 Position, Displacement, and Average Velocity, 3.3 Average and Instantaneous Acceleration, 3.6 Finding Velocity and Displacement from Acceleration, 4.5 Relative Motion in One and Two Dimensions, 8 Potential Energy and Conservation of Energy, 8.2 Conservative and Non-Conservative Forces, 8.4 Potential Energy Diagrams and Stability, 10.2 Rotation with Constant Angular Acceleration, 10.3 Relating Angular and Translational Quantities, 10.4 Moment of Inertia and Rotational Kinetic Energy, 10.8 Work and Power for Rotational Motion, 13.1 Newton’s Law of Universal Gravitation, 13.3 Gravitational Potential Energy and Total Energy, 15.3 Comparing Simple Harmonic Motion and Circular Motion, 17.4 Normal Modes of a Standing Sound Wave, 1.4 Heat Transfer, Specific Heat, and Calorimetry, 2.3 Heat Capacity and Equipartition of Energy, Chapter 3 The First Law of Thermodynamics, Chapter 4 The Second Law of Thermodynamics, 4.1 Reversible and Irreversible Processes, 4.4 Statements of the Second Law of Thermodynamics. The three forces pulling at the knot are the tension [latex]{\mathbf{\overset{\to }{T}}}_{1}[/latex] in the 5.0-cm string, the tension [latex]{\mathbf{\overset{\to }{T}}}_{2}[/latex] in the 10.0-cm string, and the weight [latex]\mathbf{\overset{\to }{w}}[/latex] of the pan holding the masses. It helps a wire-walker to maintain equilibrium. We may say that an object at rest is in equilibrium or in static equilibrium. The free-body diagram and problem-solving strategy for this special case were outlined in Newton’s Laws of Motion and Applications of Newton’s Laws. We can use the Pythagorean theorem to solve this triangle, shown in Figure 12.8, and find the sine and cosine of the angles \(\alpha_{1}\) and \(\alpha_{2}\). ∑ k F → k = 0 →. Adopted a LibreTexts for your class? From the free-body diagram, we read that torque \(\tau_{F}\) causes clockwise rotation about the pivot at CM, so its sense is negative; and torque \(\tau_{R}\) causes counterclockwise rotation about the pivot at CM, so its sense is positive. If an odd number of forces act on an object, the object cannot be in equilibrium. Since all particles in equilibrium have constant velocity, it is always possible to find an inertial reference frame in which the particle is stationary with respect to the frame. Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium: ∑k →F k = →0. 12.1 Conditions for Static Equilibrium Copyright © 2016 by OpenStax. The first equilibrium condition, Equation \ref{12.7}, reads, \[+F_{F} - w + F_{R} = 0 \ldotp \label{12.11}\], This condition is trivially satisfied because when we substitute the data, Equation \ref{12.11} becomes +0.52w − w + 0.48w = 0. However, when rotational and translational equilibrium conditions hold simultaneously in one frame of reference, then they also hold in any other inertial frame of reference, so that the net torque about any axis of rotation is still zero. If a particle in equilibrium has zero velocity, that particle is in static equilibrium. In Physics, equilibrium is the state in which all the individual forces (and torques) exerted upon an object are balanced. Find the magnitude of the tension in each supporting cable shown below. The maximum tension that the string can support is 2.80 N. Mass is added gradually to the pan until one of the strings snaps. A passenger car with a 2.5-m wheelbase has 52% of its weight on the front wheels on level ground, as illustrated in Figure. In part (a), we are asked for a distance; thus, the second condition (regarding torques) must be used, since the first (regarding only forces) has no distances in it. This vector equation is equivalent to the following three scalar equations for the components of the net force: Analogously to Figure, we can state that the rotational acceleration [latex]\mathbf{\overset{\to }{\alpha }}[/latex] of a rigid body about a fixed axis of rotation is caused by the net torque acting on the body, or. The lines of action of both normal reaction forces are perpendicular to their lever arms, so in Equation \ref{12.10}, we have |sin \(\theta\)| = 1 for both forces. At all times, the static equilibrium conditions given by Equation \ref{12.7} through Equation \ref{12.9} are satisfied. However, the second condition involves torque, which is defined as a cross product, \(\vec{\tau}_{k} = \vec{r}_{k} \times \vec{F}_{k}\), where the position vector \(\vec{r}_{k}\) with respect to the axis of rotation of the point where the force is applied enters the equation. Both push horizontally and perpendicular to the door. The CM is located somewhere between the points where the normal reaction forces act, somewhere at a distance x from the point where FR acts. Ladder problem: A uniform ladder of length L and mass Merleans against a smooth (frictionless) vertical wall. When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850 m from the hinges. When the center of gravity moves outside of the base of support, gravitational torque rotates the box in the opposite direction, and the box rolls over. Since both consumers in perfectly competitive markets are faced with the same prices the condition for joint or general equilibrium of both consumers is. Which string is it? If we set the acceleration to zero in Equation \ref{12.1}, we obtain the following equation: The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium: \[\sum_{k} \vec{F}_{k} = \vec{0} \ldotp \label{12.2}\]. This means that the net force must be zero – called translational equilibrium. According to Newton’s second law of motion, the linear acceleration of a rigid body is caused by a net force acting on it, or. [/latex], [latex]\sum _{k}{\tau }_{kx}=0,\quad \sum _{k}{\tau }_{ky}=0,\quad \sum _{k}{\tau }_{kz}=0. It is worth noting that this equation for equilibrium is generally valid for rotational equilibrium about any axis of rotation (fixed or otherwise). [/latex], [latex]\sum _{k}{{\mathbf{\overset{\to }{\tau }}}^{\prime }}_{k}=\sum _{k}{{\mathbf{\overset{\to }{r}}}^{\prime }}_{k}\times {\mathbf{\overset{\to }{F}}}_{k}=\sum _{k}({\mathbf{\overset{\to }{r}}}_{k}-\mathbf{\overset{\to }{R}})\times {\mathbf{\overset{\to }{F}}}_{k}=\sum _{k}{\mathbf{\overset{\to }{r}}}_{k}\times {\mathbf{\overset{\to }{F}}}_{k}-\sum _{k}\mathbf{\overset{\to }{R}}\times {\mathbf{\overset{\to }{F}}}_{k}=\sum _{k}{\mathbf{\overset{\to }{\tau }}}_{k}-\mathbf{\overset{\to }{R}}\times \sum _{k}{\mathbf{\overset{\to }{F}}}_{k}=\mathbf{\overset{\to }{0}}. Recall that the CM has a special physical meaning: When an external force is applied to a body at exactly its CM, the body as a whole undergoes translational motion and such a force does not cause rotation. Definition of Static Stability: In Fig. The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium: (12.2.2) ∑ k F → k = 0 →. We then obtain two equilibrium equations for the tensions: The equilibrium equation for the x-direction tells us that the tension [latex]{T}_{1}[/latex] in the 5.0-cm string is twice the tension [latex]{T}_{2}[/latex] in the 10.0-cm string. Suppose we place the axis of rotation at CM, as indicated in the free-body diagram for the car. A small pan of mass 42.0 g is supported by two strings, as shown in Figure. This means that a body in equilibrium can be moving, but if so, its linear and angular velocities must be constant. The answer is x = 0.52d = 0.52(2.5 m) = 1.3 m. Solution Choosing the pivot at the position of the front axle does not change the result. Similarly, in Equation \ref{12.7}, we assign the + sign to force components in the + x-direction and the − sign to components in the − x-direction. Another set of conditions must be met for an object to be in static equilibrium. View this demonstration to see two forces act on a rigid square in two dimensions. [/latex] The second equilibrium condition, Figure, reads, where [latex]{\tau }_{\text{F}}[/latex] is the torque of force [latex]{F}_{\text{F}},\,{\tau }_{w}[/latex] is the gravitational torque of force w, and [latex]{\tau }_{\text{R}}[/latex] is the torque of force [latex]{F}_{\text{R}}. At this point, we are ready to write the equilibrium conditions for the car. What can you say about the velocity of a moving body that is in dynamic equilibrium? However, when rotational and translational equilibrium conditions hold simultaneously in one frame of reference, then they also hold in any other inertial frame of reference, so that the net torque about any axis of rotation is still zero. Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations All examples in this chapter are planar problems. True, as the sum of forces cannot be zero in this case unless the force itself is zero. Accordingly, we use equilibrium conditions in the component form of to .We introduced a problem-solving strategy in to illustrate the physical meaning of the equilibrium conditions. As long as the elevator moves up at a constant speed, the result stays the same because the weight [latex]w[/latex] does not change. In sum, the title means that: due to the balanced forces and torques, the object will not move, neither linearly or rotationally. Thus, we identify three forces acting on the body (the car), and we can draw a free-body diagram for the extended rigid body, as shown in Figure. Suppose vector \(\vec{R}\) is the position of the origin of a new inertial frame of reference S′ in the old inertial frame of reference S. From our study of relative motion, we know that in the new frame of reference S′, the position vector \(\vec{r}′_{k}\) of the point where the force \(\vec{F}_{k}\) is applied is related to \(\vec{r}_{k}\) via the equation, \[\vec{r}'_{k} = \vec{r}_{k} - \vec{R} \ldotp\]. Now we need to decide on the location of the pivot point. [/latex] The preceding equation can be solved for the critical mass m that breaks the string: Suppose that the mechanical system considered in this example is attached to a ceiling inside an elevator going up. [latex]\sum _{k}{\mathbf{\overset{\to }{F}}}_{k}=m{\mathbf{\overset{\to }{a}}}_{\text{CM}}. 3.1.1 Conditions for Equilibrium … Static Equilibrium 3.1 The Important Stuff In this chapter we study a special case of the dynamics of rigid objects covered in the last two chapters. View this demonstration to experiment with stable and unstable positions of a box. The string breaks when the tension reaches the critical value of T1 = 2.80 N. The preceding equation can be solved for the critical mass m that breaks the string: \[m = \frac{2.5}{\sqrt{5}} \frac{T_{1}}{g} - M = \frac{2.5}{\sqrt{5}} \frac{2.80\; N}{9.8\; m/s^{2}} - 0.042\; kg = 0.277\; kg = 277.0\; g \ldotp\]. There are two conditions necessary for static equilibrium: the net force on a body equals zero and the net torque on a body equals zero. When a body in a selected inertial frame of reference neither rotates nor moves in translational motion, we say the body is in static equilibrium in this frame of reference. Accordingly, we use equilibrium conditions in the component form of Equation 12.7 to Equation 12.9. The maximum tension that the string can support is 2.80 N. Mass is added gradually to the pan until one of the strings snaps. According to Newton’s second law of motion, the linear acceleration of a rigid body is caused by a net force acting on it, or, \[\sum_{k} \vec{F}_{k} = m \vec{a}_{CM} \ldotp \label{12.1}\]. For structures in a plane, three equations of equilibrium are used for the determination of external and conditions of equilibrium of forces The first equilibrium condition, (Figure), is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws. The knot can be treated as a point; therefore, we need only the first equilibrium condition. [/latex], [latex]\sum _{k}{\mathbf{\overset{\to }{F}}}_{k}=\mathbf{\overset{\to }{0}}. With this choice of axis, the net torque has only a z-component, all forces that have non-zero torques lie in the xy-plane, and therefore contributions to the net torque come from only the x- and y-components of external forces. "Static" means stationary or at rest. Our task is to find x. [/latex] At these contact points, the car experiences normal reaction forces with magnitudes [latex]{F}_{\text{F}}=0.52w[/latex] and [latex]{F}_{\text{R}}=0.48w[/latex] on the front and rear axles, respectively. Draw a free-body diagram for a rigid body acted on by forces. In Equation \ref{12.9}, net torque is the sum of terms, with each term computed from Equation \ref{12.10}, and each term must have the correct sense. Static Equilibrium The special situation in which the net force on an object turns out to be zero, called static equilibrium, tells you immediately that the object isn’t accelerating. We say that a rigid body is in equilibrium when both its linear and angular acceleration are zero relative to an inertial frame of reference. In such a case, the object can be effectively treated like a point mass. You can vary magnitudes of the forces and their lever arms and observe the effect these changes have on the square. If the object is moving with some velocity, it will remain moving with that exact same velocity. An object in static equilibrium has zero net force acting upon it. Thus, we identify three forces acting on the body (the car), and we can draw a free-body diagram for the extended rigid body, as shown in Figure \(\PageIndex{4}\). If there is only one external force (or torque) acting on an object, it cannot be in equilibrium. Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations All examples in this chapter are planar problems. We say that a rigid body is in static equilibrium when it is at rest in our selected frame of reference. The first and second equilibrium conditions are stated in a particular reference frame. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. The fundamental and basic condition for static equilibrium is that an object must not be experiencing any type of motion, irrespective of translational or rotational. Still, in all cases the shorter string breaks first. The weight of the structure is negligible. By the end of this section, you will be able to: We say that a rigid body is in equilibrium when both its linear and angular acceleration are zero relative to an inertial frame of reference. Furthermore, an object which is in translational equilibrium does not travel from one place to another. When the CM is located off the axis of rotation, a net gravitational torque occurs on an object. The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium: ∑k →F k =→0. By setting to zero the right-hand side of Equation \ref{12.4}, we obtain the second equilibrium condition: The second equilibrium condition for the static equilibrium of a rigid body expresses rotational equilibrium: \[\sum_{k} \vec{\tau}_{k} = \vec{0} \ldotp \label{12.5}\]. condition that is called static equilibrium of an extended object. Recall that the CM has a special physical meaning: When an external force is applied to a body at exactly its CM, the body as a whole undergoes translational motion and such a force does not cause rotation. Static equilibrium and dynamic equilibrium are termed when the object is at rest and moving in a constant velocity … We also know that the car is an example of a rigid body in equilibrium whose entire weight w acts at its CM. False, provided forces add to zero as vectors then equilibrium can be achieved. Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations All examples in this chapter are planar problems. A body is in equilibrium when it remains either in uniform motion (both translational and rotational) or at rest. It is balanced when the beam remains level. We then obtain two equilibrium equations for the tensions: in y-direction, \[\frac{2T_{1}}{\sqrt{5}} + \frac{T_{2}}{\sqrt{5}} = (M + m)g \ldotp\]. Identify the physical conditions of static equilibrium. The condition Fnet =0 F net = 0 must be true for both static equilibrium, where the object’s velocity is zero, and dynamic equilibrium, where the object is moving at a constant velocity. The first equilibrium condition, (Figure), is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws. The knot can be treated as a point; therefore, we need only the first equilibrium condition. Included with Brilliant Premium Rope Statics. In this special case, we need not worry about the second equilibrium condition, Figure, because all torques are identically zero and the first equilibrium condition (for forces) is the only condition to be satisfied. When the center of gravity moves outside of the base of support, gravitational torque rotates the box in the opposite direction, and the box rolls over. How does this help? With the help of the free-body diagram, we identify the force magnitudes FR = 0.48w and FF = 0.52w, and their corresponding lever arms rR = x and rF = d − x. Because the motion is relative, what is in static equilibrium to us is in dynamic equilibrium to the moving observer, and vice versa. Numerous examples are worked through on this Tutorial page. The first and second equilibrium conditions are stated in a particular reference frame. We can choose any point as the location of the axis of rotation (z-axis). Static Equilibrium The special situation in which the net force on an object turns out to be zero, called static equilibrium, tells you immediately that the object isn’t accelerating. The free-body diagram and problem-solving strategy for this special case were outlined in Newton’s Laws of Motion and Applications of Newton’s Laws. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. At this point, we are ready to write the equilibrium conditions for the car. For planar equilibrium problems with rotation about a fixed axis, which we consider in this chapter, we can reduce the number of equations to three. The same rule must be consistently followed in Equation \ref{12.8}, when computing force components along the y-axis. The three forces pulling at the knot are the tension \(\vec{T}_{1}\) in the 5.0-cm string, the tension \(\vec{T}_{2}\) in the 10.0-cm string, and the weight \(\vec{w}\) of the pan holding the masses. This example shows that when solving static equilibrium problems, we are free to choose the pivot location. magnitude and direction of the force, and its lever arm. This video introduces the concept of static equilibrium in physics and a basic strategy to solve these static problems. A small 1000-kg SUV has a wheel base of 3.0 m. If 60% if its weight rests on the front wheels, how far behind the front wheels is the wagon’s center of mass? The free-body diagram for this pivot location is presented in Figure 12.6. Identify the physical conditions of static equilibrium. View this demonstration to see three weights that are connected by strings over pulleys and tied together in a knot. 8. [/latex] We can now write the second equilibrium condition, Figure, explicitly in terms of the unknown distance x: Here the weight w cancels and we can solve the equation for the unknown position x of the CM. You will see a typical equilibrium situation involving only the first equilibrium condition in the next example. The origin of a selected frame of reference is called the pivot point. A special case of static equilibrium occurs when all external forces on an object act at or along the axis of rotation or when the spatial extension of the object can be disregarded. Explain which one of the following situations satisfies both equilibrium conditions: (a) a tennis ball that does not spin as it travels in the air; (b) a pelican that is gliding in the air at a constant velocity at one altitude; or (c) a crankshaft in the engine of a parked car. Equilibrium is achieved, which is static equilibrium in this case. ∑ k F → k = 0 →. This gravitational torque may rotate the object if there is no support present to balance it. The magnitude of the gravitational torque depends on how far away from the pivot the CM is located. We can use the Pythagorean theorem to solve this triangle, shown in Figure, and find the sine and cosine of the angles [latex]{\alpha }_{1}[/latex] and [latex]{\alpha }_{2}. Three static properties are observed in a general equilibrium solution, reached with a free competitive market mechanism: (a) Efficient allocation of resources among firms (equilibrium of production).