12.2 Examples of Static Equilibrium _ University Physics Volume 1.pdf

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Unformatted text preview: 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 University Physics Volume 1 12 Static Equilibrium and Elasticity 12.2 Examples of Static Equilibrium -… 1/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 LEARNING OBJECTIVES By the end of this section, you will be able to: Identify and analyze static equilibrium situations Set up a free-body diagram for an extended object in static equilibrium Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of (Figure) to (Figure). We introduced a problem-solving strategy in (Figure) to illustrate the physical meaning of the equilibrium conditions. 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. Problem-Solving Strategy: Static Equilibrium 1. Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly. 2. Set up a free-body diagram for the object. (a) Choose the xy-reference frame for the problem. Draw a free-body diagram for the object, including only the forces that act on it. When suitable, represent the forces in terms of their components in the chosen reference frame. As you do this for each force, cross out the original force so that you do not erroneously include the same force twice in equations. Label all forces—you will need this for correct computations of net forces in the x– and y-directions. For an unknown -… 2/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 force, the direction must be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The correct direction is determined by the sign that you obtain in the final solution. A plus sign (+) means that the working direction is the actual direction. A minus sign (−) means that the actual direction is opposite to the assumed working direction. (b) Choose the location of the rotation axis; in other words, choose the pivot point with respect to which you will compute torques of acting forces. On the free-body diagram, indicate the location of the pivot and the lever arms of acting forces—you will need this for correct computations of torques. In the selection of the pivot, keep in mind that the pivot can be placed anywhere you wish, but the guiding principle is that the best choice will simplify as much as possible the calculation of the net torque along the rotation axis. 3. Set up the equations of equilibrium for the object. (a) Use the freebody diagram to write a correct equilibrium condition (Figure) for force components in the x-direction. (b) Use the free-body diagram to write a correct equilibrium condition (Figure) for force components in the y-direction. (c) Use the free-body diagram to write a correct equilibrium condition (Figure) for torques along the axis of rotation. Use (Figure) to evaluate torque magnitudes and senses. 4. Simplify and solve the system of equations for equilibrium to obtain unknown quantities. At this point, your work involves algebra only. Keep in mind that the number of equations must be the same as the number of unknowns. If the number of unknowns is larger than the number of equations, the problem cannot be solved. 5. Evaluate the expressions for the unknown quantities that you obtained in your solution. Your final answers should have correct numerical values and correct physical units. If they do not, then use the previous steps to track back a mistake to its origin and correct it. Also, you may independently check for your numerical answers by shifting the pivot to a different location and solving the problem again, which is what we did in (Figure). -… 3/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 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. Also note that a free-body diagram for an extended rigid body that may undergo rotational motion is different from a free-body diagram for a body that experiences only translational motion (as you saw in the chapters on Newton’s laws of motion). 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. EXAMPLE The Torque Balance Three masses are attached to a uniform meter stick, as shown in (Figure). The mass of the meter stick is 150.0 g and the masses to the left of the fulcrum are m m2 = 75.0 g. 1 = 50.0 g and Find the mass m that balances the system 3 when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced. Figure 12.9 In a torque balance, a horizontal beam is supported at a fulcrum (indicated by S) and masses are attached to both sides of the fulcrum. The system is in static equilibrium when the beam does not rotate. It is balanced when the beam remains level. -… 4/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 Strategy For the arrangement shown in the figure, we identify the following five forces acting on the meter stick: w 1 = m1 g of mass m w = mg is the weight of mass m w 2 = m2 g is the weight is the weight of the entire meter stick; w the weight of unknown mass m FS 1; 2; 3 = m3 g is 3; is the normal reaction force at the support point S. We choose a frame of reference where the direction of the yaxis is the direction of gravity, the direction of the x-axis 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. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot (Figure). 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 key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w, at the 50-cm mark. -… 5/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 Figure 12.10 Free-body diagram for the meter stick. The pivot is chosen at the support point S. Solution With (Figure) and (Figure) for reference, we begin by finding the lever arms of the five forces acting on the stick: r1 = 30.0 cm + 40.0 cm = 70.0 cm r2 = 40.0 cm r = 50.0 cm − 30.0 cm = 20.0 cm rS = 0.0 cm (because FS is attached at the pivot) r3 = 30.0 cm. Now we can find the five torques with respect to the chosen pivot: τ1 = +r1 w1 sin 90° = +r1 m1 g (counterclockwise rotation τ2 = +r2 w2 sin 90° = +r2 m2 g (counterclockwise rotation τ = +rw sin 90° = +rmg (gravitational torque) τS = rS FS sin θS = 0 (because rS = 0 cm) τ3 = −r3 w3 sin 90° = −r3 m3 g (clockwise rotation, negat The second equilibrium condition (equation for the torques) for the meter stick is τ1 + τ2 + τ + τS + τ3 = 0. When substituting torque values into this equation, we can -… 6/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 omit the torques giving zero contributions. In this way the second equilibrium condition is +r1 m1 g + r2 m2 g + rmg − r3 m3 g = 0. Selecting the +y-direction to be parallel to F→ S, the first equi- librium condition for the stick is −w1 − w2 − w + FS − w3 = 0. Substituting the forces, the first equilibrium condition becomes −m1 g − m2 g − mg + FS − m3 g = 0. We solve these equations simultaneously for the unknown values m and F 3 S. In (Figure), we cancel the g factor and re- arrange the terms to obtain r3 m3 = r1 m1 + r2 m2 + rm. To obtain m we divide both sides by r , so we have 3 m3 = = r1 r3 70 30 m1 + 3 r2 r3 m2 + (50.0 g) + 40 30 r r3 m (75.0 g) + 20 30 (150.0 g) = 316.0 2 3 g ≃ 317 To find the normal reaction force, we rearrange the terms in (Figure), converting grams to kilograms: FS = (m1 + m2 + m + m3 )g = (50.0 + 75.0 + 150.0 + 316.7) × 10 −3 kg × 9.8 m s 2 = 5.8 N Significance Notice that (Figure) is independent of the value of g. The torque balance may therefore be used to measure mass, since variations in g-values on Earth’s surface do not affect these measurements. This is not the case for a spring balance because it measures the force. CHECK YOUR UNDERSTANDING Repeat (Figure) using the left end of the meter stick to cal- -… 7/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 culate the torques; that is, by placing the pivot at the left end of the meter stick. Show Solution In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by (Figure) and (Figure). 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. EXAMPLE Forces in the Forearm A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in (Figure). His forearm is positioned at β = 60° with respect to his upper arm. 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? Assume that the forearm’s weight is negligible. Give your final answers in SI units. -… 8/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 Figure 12.11 The forearm is rotated around the elbow (E) by a contraction of the biceps muscle, which causes tension T→ M. Strategy We identify three forces acting on the forearm: the unknown force F→ at the elbow; the unknown tension T→ M in the muscle; → with magnitude w = 50 lb. We adopt the and the weight w 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 x-axis makes an angle β = 60° with the vertical. The y- axis is perpendicular to the x-axis. Now we set up the freebody diagram for the forearm. First, we draw the axes, the pivot, and the three vectors representing the three identified forces. Then we locate the angle β and represent each force by its x– and y-components, remembering to cross out the original force vector to avoid double counting. Finally, we label the forces and their lever arms. The free-body diagram for the forearm is shown in (Figure). 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 -… 9/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 first equilibrium condition, one equation for each component of the net force acting on the forearm. Figure 12.12 Free-body diagram for the forearm: The pivot is located at point E (elbow). Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the ycomponents of the forces because the x-components of the forces are all parallel to their lever arms, so that for any of them we have sin θ = 0 in (Figure). For the y-components we have θ = ±90° in (Figure). 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 and of w y y. Solution We see from the free-body diagram that the x-component of the net force satisfies the equation +Fx + Tx − wx = 0 and the y-component of the net force satisfies +Fy + Ty − wy = 0. (Figure) and (Figure) are two equations of the first equilibrium condition (for forces). Next, we read from the free-body diagram that the net torque along the axis of rotation is … 10/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 +rT Ty − rw wy = 0. (Figure) is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are r = 1.5 in. T and r w = 13.0 in. At this point, we do not need to convert inches into SI units, because as long as these units are consistent in (Figure), they cancel out. Using the freebody diagram again, we find the magnitudes of the component forces: Fx = F cos β = F cos 60° = F / 2 Tx = T cos β = T cos 60° = T / 2 wx = w cos β = w cos 60° = w / 2 Fy = F sin β = F sin 60° = F √3 / 2 Ty = T sin β = T sin 60° = T √3 / 2 wy = w sin β = w sin 60° = w√3 / 2. We substitute these magnitudes into (Figure), (Figure), and (Figure) to obtain, respectively, F /2 + T /2 − w/2 = 0 F √3 / 2 + T √3 / 2 − w√3 / 2 = 0 rT T √ 3 / 2 − rw w √ 3 / 2 = 0. When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T, because (Figure) for the x-component is equivalent to (Figure) for the y-component. In this way, we obtain the first equilibrium condition for forces F + T − w = 0 and the second equilibrium condition for torques rT T − rw w = 0. The magnitude of tension in the muscle is obtained by solving (Figure): T = rw rT w = 13.0 1.5 (50 lb) = 433 1 3 lb ≃ 433.3 lb. The force at the elbow is obtained by solving (Figure): F = w − T = 50.0 lb − 433.3 lb = −383.3 lb. … 11/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 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 F = 383.3 lb = 383.3(4.448 N) = 1705 N downward T = 433.3 lb = 433.3(4.448 N) = 1927 N upward. Significance 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. Solution Suppose we adopt a reference frame with the direction of the y-axis along the 50-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). We draw the free-body diagram for the forearm as shown in (Figure), indicating the pivot, the acting forces and their lever arms with respect to the pivot, and the angles θ and θ that the forces T→ w M T → (respectively) make with their and w lever arms. In the definition of torque given by (Figure), the angle θ is the direction angle of the vector T→ T M, counted coun- terclockwise from the radial direction of the lever arm that always points away from the pivot. By the same convention, the angle θ is measured counterclockwise from the radial direcw → Done this way, the nontion of the lever arm to the vector w. zero torques are most easily computed by directly substituting into (Figure) as follows: … 12/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 τT = rT T sin θT = rT T sin β = rT T sin 60° = +rT T √3 / 2 τw = rw w sin θw = rw w sin(β + 180°) = −rw w sin β = −rw w√3 / Figure 12.13 Free-body diagram for the forearm for the equivalent solution. The pivot is located at point E (elbow). The second equilibrium condition, τ T + τw = 0, can be now written as rT T √3 / 2 − rw w√3 / 2 = 0. From the free-body diagram, the first equilibrium condition (for forces) is −F + T − w = 0. (Figure) is identical to (Figure) and gives the result T = 433.3 lb. (Figure) gives F = T − w = 433.3 lb − 50.0 lb = 383.3 lb. 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. CHECK YOUR UNDERSTANDING … 13/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 Repeat (Figure) assuming that the forearm is an object of uniform density that weighs 8.896 N. Show Solution EXAMPLE A Ladder Resting Against a Wall A uniform ladder is L = 5.0 m long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in (Figure). The inclination angle between the ladder and the rough floor is β = 53°. Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction μ s at the interface of the ladder with the floor that prevents the ladder from slipping. Figure 12.14 A 5.0-m-long ladder rests against a frictionless wall. … 14/33 8/31/2021 12.2 Examples of Static Equilibrium | University Physics Volume 1 Strategy 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. The second force is the static friction force f = μ N s directed horizontally along the floor toward the wall —this force prevents the ladder from slipping. 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 pivo...
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