File Name: statics of rigid bodies problems and solutions .zip
- Engineering mechanics solved problems pdf
- 12.3: Examples of Static Equilibrium
- 2 Rigid Bodies Problem Solutions
- Engineering mechanics solved problems pdf
All examples in this chapter are planar problems.
Statics is the study of bodies and structures that are in equilibrium. For a body to be in equilibrium , there must be no net force acting on it. In addition, there must be no net torque acting on it.
Engineering mechanics solved problems pdf
We set up an coordinate system to specify where the object is by the position and its posture by the orientation. Often, we are more interested in the change of physical quantities with respect to time, e. The rotation effect is insignificant because it is just a point.
Whether the body can be treated as the particle or not depends on the relative dimensions in the problem and how much detailed of the solution we are interested in. For example the motion of an ingot can be analyzed by assuming the object being rigid.
Knowledge of the mechanics of the deformable material must be used along with Dynamics in order to determine the absolute motion of the nonrigid bodies. Let us consider some examples to see the difference of each term. If we have an object and consider the very small substance of the body. For differential element analysis of the body, the small substance can be treated as a particle. However, the substance must be handled as connecting objects had the molecular effects in the body are of concern.
Or think of an airplane. Even of its huge size, the whole airplane may be modeled as a point in flight speed analysis along the route. But if the rotational motion, such as yawing or pitching, of the airplane body is important, its size does matter. The next two examples are to show whether an object is considered rigid or nonrigid depends on how much detailed of the problem we would like to analyze.
Truss can just be looked as a rigid body for the preliminary design of truss structure. But we must think of the truss elasticity if we were to choose the material for that truss. A stiff linkage of the robot may be considered a rigid body.
However, the n-connecting linkages, treated as a whole, to form the robot arm is an example of nonrigid body. Note the body-fixed inertia of the nonrigid body is not constant. Particularly, the vector term is explained more in length because it is fundamental to many dynamical variables. Time, volume, density, speed, energy, and mass are some examples. Examples are displacement, velocity, acceleration, force, moment, and momentum. Vectors cand be classified into 3 types: free vector, sliding vector, and fixed vector.
That is, only its magnitude and direction do matter. Some examples are the displacement vector of a pure translational rigid object, or the couple vector of a rigid body. Free vector is free to slide and translate as long as its direction and magnitude are maintained.
In other words, its line of action and point of application do not matter. External force or moment acting on the rigid body falls under this category. Therefore sliding vector has a freedom to slide along the fixed line of action. External force or moment acting onto the nonrigid body must be dealt with as the fixed vector due to the deformable effect of the object.
Representation of VectorsThere are many notations to represent a vector quantity, i. If we would like to tell only the magnitude, v or v may be used. Keep in mind that the complete representation of a vector must be able to determine its magnitude, direction, line of action, and point of application. See Fig. In the algebraic approach, cosine law and sine law are used in determining the magnitude and direction of the resultant vector from the addition of two vectors, as shown in Fig.
The magnitude, i. Different coordinate systems can be defined and used to solve the same problem because the vector quantities are invariant to the coordinate systems.
However, some of them will be more appropriate to the problem at hand than others. Usually we get used to the coordinate systems of which their coordinate axes are perpendicular. They are called rectangular coordinate systems. In some situation, non-rectangular coordinate system may be needed. After we set up the coordinate system, the vector can be described by its components along the coordinate axis directions.
As seen in Fig. Vector components are the adjacent sides of the parallelogram. Therefore, cosine and sine laws can be used to determine the components. In case of rectangular coordinate system, vector components are determined simply by the dot product of the vector with unit vector along those axes. In other words, the components of the vector associated with the given rectangular coordinate system are the orthogonal projection of the vector onto the corresponding coordinate axes.
The components in this special case is the orthogonal projection of the vector onto the corresponding coordinate axes, which can be calculated from the dot product of the vector with the respective unit vector. It can be concluded from the equation that the three angles are dependent. Only two of them are enough to specify the orientation of the vector.
Newton's LawsIn this section, we briefly mention the Newton's laws that describe the motion of the particle under low velocity. The first law states:"A particle remains at rest or continue to move in a straight line with a uniform velocity if there is no unbalanced force acting on it. Newton's third law states:"The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear.
The rectangular coordinate system is set up as a mean to describe the force system. Force and their consequences, moment and couple, can then be described using the defined coordinate system. Finally, resultant force system will be determined as the simplest representation of the complex system of forces, moments, and couples.
In this section, we roughly explain the concepts and basic terminologies involving with force. It is a fixed vector. For the rigid body problems, or if only the external effects of the external force onto the objects are of interested, that force can be treated as a sliding vector.
Hence the problem can make use of the principle of transmissibility. Specification of the magnitude, direction, and line of action can completely describe the force vector.
If the body is rigid enough , the force pushing at point A will generate the same resulting motion as the force pulling the object at point B. Body force Contact force occurs from the contact between the bodies, which is very common since the phenomena can be easily observed.
However, there is force between the bodies even though they are not in contact. It is the attracting force, or body force, trying to pull the bodies together, of which its magnitude is governed by the gravitational law. Distributed force Most of the forces are naturally the distributed force, which means the force acts over some area or surface such as the force developed between the tire and the road.
Concentrated force is the ideal situation that the surface or volume where the force acts on is negligible. This simplifies the situation and makes possible the preliminary analysis of the problem. In some circumstances, where the body can be assumed rigid, the distributed force can be reduced to a single resultant force.
In other words, we find the equivalent concentrated force to the original distributed force system. Its principle is to infer the force from the deformation of an elastic element. Calibration of the force sensor to the known load is necessary. Reaction force They are pair of forces expressing the interaction between bodies. They will be revealed when isolate the surrounding objects from the system of interest.
Free body diagram FBD is used to help indentify the action and reaction forces. The line of action of the combined force will be corrected. Again, the parallelogram law or the principle of moment, including the principle of transmissibility are used to obtain the correct force components. It is calculated by the dot product of the vector and the unit vector in that direction. The components of a vector, however, are usually not the same as the orthogonal projection onto the same coordinate system.
Exception is the orthogonal rectangular coordinate system. Figure 2. With the parallelogram law, sum of the components must equal to the original vector.
The trick is to make them intersect by adding an arbitrary force vector to one of the original force, and then subtracting the opposite from the other one of the original force. The new pair of forces is now unparalleled and can be added up to get the resultant force with the location of the line of action.
Solution: Use the parallelogram law and the cosine law to determine the non-orthogonal components. From the force vector addition and by the cosine law, Solution: The indicated force P is the force done by the robot on the cylindrical part.
We set up the coordinate frame n 1 t 1 and n 2 t 2 where their axes are perpendicular and parallel to the link AB and BC, respectively. The moment is always associated with a specified point, meaning that we must specify the point in determining the moment about that point.
12.3: Examples of Static Equilibrium
For an rigid body in static equilibrium, that is a non-deformable body where forces are not concurrent, the sum of both the forces and the moments acting on the body must be equal to zero. The addition of moments as opposed to particles where we only looked at the forces adds another set of possible equilibrium equations, allowing us to solve for more unknowns as compared to particle problems. Moments, like forces, are vectors. This means that our vector equation needs to be broken down into scalar components before we can solve the equilibrium equations. In a two dimensional problem, the body can only have clockwise or counter clockwise rotation corresponding to rotations about the z axis. This means that a rigid body in a two dimensional problem has three possible equilibrium equations; that is, the sum of force components in the x and y directions, and the moments about the z axis.
2 Rigid Bodies Problem Solutions
Marghitu, Wiggins G, phone: , office hours TR: p. Hibbeler, Prentice Hall, Engineering Mechanics, Volume 1 - Statics, by J. Meriam and L.
The hammer in the figure is placed over a block of wood of 40 mm of thickness, to facilitate the extraction of the nail. If a force of N perpendicular to the hammer is required to extract the nail, find the force on the nail and the force at point A while the nail starts to be removed in the position shown. Assume that the weight of the hammer can be neglected, compared to the other forces and that there is enough friction in A to prevent the hammer from sliding.
Engineering mechanics solved problems pdf
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All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Equation We introduced a problem-solving strategy in Example Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. We proceed in five practical steps. Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium.
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