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Equilibrium Physics Problems And Solutions Pdf

equilibrium physics problems and solutions pdf

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Statics Problems And Solutions Pdf

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. In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone.

The reason for this is that in analyzing rotation, we must identify torques acting on the body, and torque depends both on the acting force and on its lever arm. Here, the free-body diagram for an extended rigid body helps us identify external torques.

The mass of the meter stick is Find the mass m3 that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced.

For the arrangement shown in the figure, we identify the following five forces acting on the meter stick:. We choose a frame of reference where the direction of the y-axis is the direction of gravity, the direction of the xaxis is along the meter stick, and the axis of rotation the z-axis is perpendicular to the x-axis and passes through the support point S. In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates.

Now we are ready to set up the free-body diagram for the meter stick. At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The second equilibrium condition equation for the torques for the meter stick is.

When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is. We solve these equations simultaneously for the unknown values m 3 and F S. This is not the case for a spring balance because it measures the force. Repeat Example In the next example, we show how to use the first equilibrium condition equation for forces in the vector form given by Equation We present this solution to illustrate the importance of a suitable choice of reference frame.

Although all inertial reference frames are equivalent and numerical solutions obtained in one frame are the same as in any other, an unsuitable choice of reference frame can make the solution quite lengthy and convoluted, whereas a wise choice of reference frame makes the solution straightforward.

We show this in the equivalent solution to the same problem. This particular example illustrates an application of static equilibrium to biomechanics. A weightlifter is holding a The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow.

Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Give your final answers in SI units. We adopt the frame of reference with the x-axis along the forearm and the pivot at the elbow.

The vertical direction is the direction of the weight, which is the same as the direction of the upper arm. The y-axis is perpendicular to the x-axis. Now we set up the free-body diagram for the forearm. First, we draw the axes, the pivot, and the three vectors representing the three identified forces.

Finally, we label the forces and their lever arms. At this point, we are ready to set up equilibrium conditions for the forearm. Each force has x- and y-components; therefore, we have two equations for the first equilibrium condition, one equation for each component of the net force acting on the forearm.

Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of T y and of w y. We see from the free-body diagram that the x-component of the net force satisfies the equation.

Next, we read from the free-body diagram that the net torque along the axis of rotation is. At this point, we do not need to convert inches into SI units, because as long as these units are consistent in Equation Using the free-body diagram again, we find the magnitudes of the component forces:. In this way, we obtain the first equilibrium condition for forces. The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram.

In the final answer, we convert the forces into SI units of force. The answer is. Two important issues here are worth noting. The first concerns conversion into SI units, which can be done at the very end of the solution as long as we keep consistency in units. The second important issue concerns the hinge joints such as the elbow. In the initial analysis of a problem, hinge joints should always be assumed to exert a force in an arbitrary direction , and then you must solve for all components of a hinge force independently.

In this example, the elbow force happens to be vertical because the problem assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a general rule. Suppose we adopt a reference frame with the direction of the y-axis along the lb weight and the pivot placed at the elbow.

In this frame, all three forces have only y-components, so we have only one equation for the first equilibrium condition for forces. In the definition of torque given by Equation Done this way, the non-zero torques are most easily computed by directly substituting into Equation We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components.

We can identify four forces acting on the ladder. The first force is the normal reaction force N from the floor in the upward vertical direction. These two forces act on the ladder at its contact point with the floor. The third force is the weight w of the ladder, attached at its CM located midway between its ends.

The fourth force is the normal reaction force F from the wall in the horizontal direction away from the wall, attached at the contact point with the wall.

There are no other forces because the wall is slippery, which means there is no friction between the wall and the ladder. Based on this analysis, we adopt the frame of reference with the y-axis in the vertical direction parallel to the wall and the x-axis in the horizontal direction parallel to the floor. In this frame, each force has either a horizontal component or a vertical component but not both, which simplifies the solution.

We select the pivot at the contact point with the floor. With our choice of the pivot location, there is no torque either from the normal reaction force N or from the static friction f because they both act at the pivot.

With the help of the free-body diagram, we identify the angles to be used in Equation Now we are ready to use Equation The net force on the ladder at the contact point with the floor is the vector sum of the normal reaction from the floor and the static friction forces:. We should emphasize here two general observations of practical use. First, notice that when we choose a pivot point, there is no expectation that the system will actually pivot around the chosen point.

The ladder in this example is not rotating at all but firmly stands on the floor; nonetheless, its contact point with the floor is a good choice for the pivot. Second, notice when we use Equation As long as the angle in Equation This is because torque is the vector product of the lever-arm vector crossed with the force vector, and Equation This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping.

For the situation described in Example Find the forces on the hinges when the door rests half-open. The forces that the door exerts on its hinges can be found by simply reversing the directions of the forces that the hinges exert on the door. Hence, our task is to find the forces from the hinges on the door.

The CM is located at the geometrical center of the door because the slab has a uniform mass density. In the free-body diagram, we represent the two forces at the hinges by their vector components, whose assumed orientations are arbitrary. Because there are four unknowns A x , B x , A y , and B y , we must set up four independent equations. One equation is the equilibrium condition for forces in the x-direction.

The second equation is the equilibrium condition for forces in the y-direction. The third equation is the equilibrium condition for torques in rotation about a hinge. Finally, we solve the equations for the unknown force components and find the forces.

From the free-body diagram for the door we have the first equilibrium condition for forces:. We select the pivot at point P upper hinge, per the free-body diagram and write the second equilibrium condition for torques in rotation about point P:. The forces on the door are.

12.3: Examples of Static Equilibrium

If an object is at equilibrium, then the forces are balanced. Balanced is the key word that is used to describe equilibrium situations. This extends from Newton's first law of motion. An object at equilibrium is either This too extends from Newton's first law of motion. If an object is at rest and is in a state of equilibrium, then we would say that the object is at "static equilibrium.

equilibrium physics problems and solutions pdf

Section Summary

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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

Equilibrium and Statics

It is one thing to have a system in equilibrium; it is quite another for it to be stable. There are three types of equilibrium : stable , unstable , and neutral. Figures throughout this module illustrate various examples.

Equilibrium Problems and Applications develops a unified variational approach to deal with single-valued, set-valued and quasi-equilibrium problems. This abstract approach is based on tools from various fields, including set-valued analysis, variational and hemivariational inequalities, fixed point theory, and optimization. Applications include models from mathematical economics, Nash equilibrium of non-cooperative games, and Browder variational inclusions. The content is self-contained and the book is mainly addressed to researchers in mathematics, economics and mathematical physics as well as to graduate students in applied nonlinear analysis. Researchers in applied mathematics and econometrics, including graduate students in applied nonlinear analysis. Researchers in mathematical physics.

Equilibrium is a special case in mechanics where all the forces acting on a body equal zero. This type of problem pops up in many situations and is important in engineering and physics. This equilibrium example problem illustrates how to determine the different forces acting on a system of forces acting on a body in equilibrium. Example Problem: A block of weight w is suspended from a rope tied to two other ropes at point O. One rope is horizontally attached to a wall and the other is fastened to the ceiling.

Contributors and Attributions

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.

The second condition necessary to achieve equilibrium involves avoiding accelerated rotation maintaining a constant angular velocity. A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges. Several familiar factors determine how effective you are in opening the door. See Figure 1.

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Equilibrium Problems and Applications

Ему хотелось домой. Он посмотрел на дверь с номером 301.

1 Comments

  1. Rinaldo P.

    02.06.2021 at 00:38
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    Problem 1: Static Equilibrium: Steel Beam and Cable In this problem you will express your answers symbolically. SC Physics I: Classical Mechanics.

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