Moment of Intertia 8. Principles of Friction 9. Applications of Friction Principles of Lifting Machines Simple Lifting Machine Support Reactions Analysis of Perfect Frames Analytical Method Analysis of Perfect Frames Graphical Method Equilibrium of strings Virtual Work Linear Motion Motion Under Variable Acceleration Relative Velocity Projectiles Motion of Rotation The last example reveals the effectiveness of hiding unwanted unknowns, the cable tension in this case, inside the boundary of the FBD.
Example 3. Solution: By following the guideline given above, the complete FBDs as shown in fig. Since the object is in equilibrium, the resultant must be nulled and hence the zero value on the right hand side. There are some things needed to be mentioned in applying the equations. The x-y coordinate system and the moment point O can be chosen arbi- trarily. Complete equilibrium in 2-D motion must satisfy all three equations.
How- ever, they are independent to each other. That is, equilibrium may only be satisfied in some generalized coordinates.
System in equilibrium may stay still or move with constant velocity. In both cases, the acceleration is zero. At this point, the reader is armed with enough information to start practicing the equilibrium problems. However it is helpful to know the characteristics of some special cases of equilibrium problems. If we classify the problem according to the force system, the following cases of equilibrium in two dimension, fig.
The other equations perpendicular force and moment equations are satisfied automatically and hence contribute nothing in solving the problems. For the concurrent force system, the moment at the point of concurrency is always satisfied. Therefore only the force equilibrium equations are usable. In the case of parallel forces, the force in the perpendicular direciton is null. Hence only the force equilibrium equation in the parallel direction and the moment equation are active.
Note that in general case of force system, all three equilibrium equations are valid. Two-force member is a body under the action of two force only. The equilibrium condition of the two-force member requires the acting forces be equal, opposite, and collinear.
Only one equation of force along its direction is then effective. One common assumption of the two-force member is that the weight of the member is negligible.
Sometimes it is the trade-off between the simplification in solving the equilibrium problem and the accuracy of the answer. Three-force member is a body under the action of three forces only. The equilibrium condition of the three-force member requires the lines of action of the three forces be concurrent. The only exception is when the three forces are parallel. In this case, two equations of forces are effective. The moment equation is satisfied automatically.
The equivalent requirement is the closure of the polygon of forces. Equation 3. Here we represent another two sets of equations which also guarantee the equi- librium of an object.
However all three requirements will be satisfied only if the resultant is zero, otherwise the moment equations will make the nonzero resultant break the force equation. However all three requirements will be satisfied only if the resultant is zero, otherwise the last moment equation will make the nonzero resultant break the first two equations.
Constraints and Statical Determinacy The equilibrium equation may not always solve all unknowns in the problem.
This is because the equilibrium condition do not provide enough equations. Simply put, if the number of un- knowns including geometrical variables is greater than the number of equations, then we cannot solve it. This is because the system has more constraints than necessary to maintain the equilibrium. This is called statically indeterminate system. Extra equations obtained from force-deformation material properties must also be applied to solve for the redundant constraints.
If the number of unknown is equal or less than the number of equilibrium equations, the system is statically determinate.
With some wrong installation of the supports, the number of active constraints may be reduced unintentionally. See case b and c of fig. Then count the number of unknowns and the number of available independent equations. If the number of unknowns is greater than the number of equations, the problem cannot be solved solely by the equilibrium conditions.
Determine the isolated system and draw the FBD. Assign a convenient set of coordinate systems. Choose suitable moment centers for calculation. Write down the governing equations, e. Choose the suitable method in solving the problem; whether it be scalar, vector, or geometric approach.
If the load-cell reading is N, determine the vertical tensile force F generated by the triceps muscle. The mass of the lower arm is 1. State any assumptions. Solution: Let us choose the system to be the lower arm. The FBD of the system is shown in fig. From the figure, there are two unknowns. Hence two equations are required, which are one moment and one force equations. First let us take the moment about O to eliminate the compressive force. Two cables are attached to the cart — one for each hand.
If the hands are together so that the cables are parallel and if each cable lies essentially in a vertical plane, determine the force P which each hand must exert on its cable in order to maintain an equilibrium position. Solution: Assume the cart is lightweight and the rail friction is negligible.
Then we select the machine and the man as our system and draw the FBD as shown in fig. It is seen that there are two unknowns which are solvable using two force equations. Only the force exerted by each hand, P, is the unknown. Solution: Write the FBD of the system which consists of the ring and the eccentric mass.
The force acting on the system are the weights, the friction force, and the normal force. Here the friction is not negligible and indeed makes the system be in equilibrium.
First we should take the moment equation about O to eliminate the unknown N. Note that the normal force does not appear in the equation. If a moment of 80 Nm is required to turn the mm diameter collar about its center O under the action of the applied force P, determine the contact force R on the smooth surface at A.
Engagement of the pin at B may be considered to occur at the periphery of the collar. Solution: First select the system of the wrench and the collar as shown in fig. Note that this is a three-force member. The moment equilibrium at B suppress the reaction R to appear in the equation. Hence three reaction forces must meet at a common point, which helps in drawing the correct forces shown in fig.
However, before setting up the equilibrium equations, let us consider the geometry at BCDO to determine the moment arm of the hydraulic force. Referring to fig. Start with the moment equation at O to eliminate unknown reaction at O. Also find the total normal reaction NB under the rear pair of wheels at B. Solution: The inclined coordinate system is suitable for this problem.
We draw the FBD of the truck as our system. The forces are its weight, the pulling load P, the friction force, and the normal force on the wheels. Note that the reaction at the front and rear wheels are different due to the incline and the center of gravity is not at the midway between the wheels. The problem states the driving force is 80 percent of the normal force of that wheel.
Hence there is only one unknown for each wheel. Totally there are 3 unknowns; the normal forces on the wheels and the pulling force. Let us apply the force equilibrium equations first. Another equation from the moment equilibrium must be solved simultaneously.
Let us choose to take the moment about A. The tension in the lower side of the belt is N. The driving motor B has a mass of kg and rotates clockwise. Determine the magnitude R of the force on the supporting pin at O. Solution: To determine the reaction at O, we must know the reaction at the spring support and the upper side tension.
Since the pulley A delivers Nm torque to a pump, there must be the moment reaction resisting the clockwise rotation of it. The moment equilibrium is then used to determine the tension force, T. Determining three unknowns are relatively straightforward. The center of buoyancy is the point through which the resultant of the buoyant force passes. Solution: When the boat is free floating no thrust or tension , the buoy- ancy force is equal to the weight and acting at the C.
At the times the boat moves backward, the anchor chain is pulled against the thrust force. This changes the buoyancy force, both in magnitude and point of application, to maintain the equilibrium. Apply the equilibrium equations to solve for three unknowns; A, B, and b. One of the rollers at B is a gear which meshes with a ring of gear teeth on the sector so as to turn the sector about its geometric center O.
Solution: The problem gives two different postures that are in equilibrium and ask for the locating the C. Therefore two equations coming from each posture of equilibrium must be constituted and solved. Hence the reaction force at the meshing teeths is as shown in fig. To avoid determining the unnecessary unknowns, NA and NB , we should use the moment equation taken at the center point O. FBD will give the answer of the left hand side of the equations, which are the resultant of the force system acting on the body of interest.
The x-y-z coordinate system and the moment point O can be chosen arbi- trarily. Complete equilibrium in 3-D motion must satisfy all six equations. Beware these are not the FBDs. Reaction force of the member in contact with the smooth surface, or ball- supported member, is normal to the surface and directed toward the mem- ber. It is usually denoted by N. Reaction force of the member in contact with the rough surface has both the normal and tangential components.
To determine these forces, we first seek for the contact plane which is tangent to both contact surfaces. Normal force, N, acts in the direction normal to the plane while tangential force, F, lies in the plane.
Tangential force is commonly recognized as the friction force. Usually there will be additional reaction force corresponding to the addi- tional constraint. In the example, the lateral force P is introduced as the rail also prevents the wheel from the sideway motion, in addition to the normal force N.
The ball and socket joint constraints the point on each linkage to always be together. This requires the joint to support a force R. If the joint is welded or completely embedded, two linkages cannot be moved relative to each other.
Therefore the additional supporting moments must be ensured to prevent the rotational motion. The thrust bearing support prevents the shaft from moving and rotating in all direction except the rotational motion about the shaft axis. Hence three dimensional reaction forces, Rx , Ry , and Rz , must be supplied.
In addition, the resisting moment in x and z- direction must be provided as well. If we classify the problem according to the force system, the following cases of equilibrium in three dimension, fig. If the force system has all forces being concurrent with a line, i. Therefore only five equilibrium equations are effective.
In the case of parallel forces, the force in the perpendicular direction is null. Hence only the force equilibrium equation in the parallel direction and the moment equations about the lines in the plane perpendicular to the parallel forces are active.
Note that in general case of force system, all six equilibrium equations are valid. Determine the reactions at O and the cable tensions. Solution: The FBD of this problem is shown in fig. There are six unknowns; three reaction force components at O and three cable tensions.
We can simply set up three force equilibrium equations and three moment equilibrium equations about any point. However solving the resulting equations may be intractable. A little thought before writing down the solution can be helpful. For this problem, the moment equilibrium about line OB suppresses five unknowns to show up. Only TAC will appear in the equation, which is easy to solve.
Examining the FBD in this manner helps avoid solving simultaneous equations unnecessarily. The sum of the forces acting on the system in any direction must be zero. The door is maintained in a vertical plane by the floor-mounted guide roller C, which bears against the bottom edge. For the position shown compute the horizontal side thrust on each of the wheels A and B, which must be accounted for in the design of the brackets.
Hence the side thrusts on the wheels must point in the opposite direction to balance the force. Solution: First the FBD will be drawn according to the given condition that the reaction on each side of the groove must equal the force supported by the end plate. However, it is rather difficult to draw and visualize the FBD of the sphere in three dimensions.
Therefore the FBD will be drawn in two orthogonal views; along and perpendicular to the V-groove. Forces acting on the sphere will then be the projected components onto that view. Also the coordinate frame x-y-z is set at the center of the sphere for convenience. If the weight of the door is supported entirely by the lower hinge A, calculate the magnitude of the total force supported by the hinge at B. Solution: From the problem statement, it is implied that there is no verti- cal support force at the upper hinge B.
This assumption must have been enforced otherwise it cannot be solved by just the equilibrium conditions. The FBD is straightforward with the five unknown supports revealed and the appropriate coordinate frame set. The mass of the lander is kg. Compute the force in each leg when the lander is resting on a horizontal surface on Mars. Solution: From the problem statement, it can be deduced that each land- ing pad supports kg force.
The leg with the landing pad is sectioned out and its FBD is drawn. Forces in each strut are then revealed. Note that the forces are all concurrent at the center of ball and socket. Hence the available equilibrium conditions are reduced to three only. Calculate the magnitude of the force supported by bearing B during application of the Nm couple to the shaft. The cable from C to D prevents the plate and shaft from turning, and the weight of the assembly is carried entirely by bearing A.
Solution: The plate and the attached vertical shaft are selected as the system and its FBD is drawn. By the statement that the weight of the assembly is carried entirely by bearing A, bearing B will experience only the radial forces.
The convenient coordinate frame is defined and now it is ready to set up the equilibrium conditions. Calculate the compression FB in the strut and the force supported by the hinge D normal to the hinge axis.
Assume that the hinges act at the extreme ends of the lower edge. Solution: The trap door is isolated and its FBD is drawn. There are the reactions at C and D, the strut force, and its weight. Reactions at D are decomposed along the coordinate frame axes. It is unnecessary to do so for the reactions at C, however. Hinge A can support thrust along the hinge axis AB, whereas hinge B supports force normal to the hinge axis only.
Note that there is no reaction component along the hinge axis at B. Only Bz then ap- pears in the equation. The collar D fastened to the shaft prevents downward motion of the shaft in its bearing. Calculate the bending moment M, the compression P , and the shear force V in the shaft at section B.
Solution: The shaft is sectioned at B fig. Reactions are the compressive force, the shear force, and the bending moment. However the shaft is free to rotate about its axis. Hence no supporting moment in this direction. Solution: The reel is supported by the structure that is connected to the reel axis. The turning force can be determined from the FBD of the reel only left of fig.
The reaction forces and the bending moments are exposed. Since these forces are not of interest, take moment about the y-axis. Note of the friction forces which are not negligible. The shaft is subjected to a torque couple of Nm, and the drum is prevented from rotating by the cord wrapped securely around it and attached to point C. Calculate the magnitudes of the forces supported by bearings A and B.
Solution: Select the drum and shaft unit and draw the FBD. Each bear- ing supports the shaft by the radial force which is projected into the x- and z- direction as shown in fig. First, take moment about B along the y-axis. Solution: By the double U-joint mechanism, O can provide support forces in any direction.
However, it can only support the moment in the direction normal to the back plate. The FBD of the boom system is shown in fig. To the mutual contacting surface, these forces have their components both in the tangential and normal directions. Force com- ponent in the tangential direction is known as the friction force. Whenever a tendency exists for one contacting surface to slide along another sur- face, the developed friction force is always in the direction opposing this tendency.
In some systems, friction is undesirable because it just plainly changes the system characteristics from the required behavior. In particular, where the slid- ing motion between parts occurs, the developed friction force results in a loss of energy. However, in many cases, friction instead functions the systems.
For example, many mechanisms employ friction as an extra force to retain their equi- librium states. A block of mass m slides on a planar surface with the applied pulling force P , as shown in fig. To begin the analysis, the free body diagram is drawn, depicted in fig.
The reaction force R the ground exerted on the block is decomposed into the normal and tangential components, according to the virtual tangent plane at the contact.
This angle can be determined approximately by the experiment. The value depends largely on the materials of two contacting surface. Local geometry of the contact point determines each infinitesimal reaction force which adds up vectorially to give the gross reaction force R.
Relationship of the friction force F and the applied force P is plotted in fig. There are three regimes of the friction force development. During this phase, the block does not move at all. This can be understood by determining the resulting force acting along the tangential direction. The force is zero which indicates the equilibrium status and hence no motion of the block.
By comparing eq. At this point, the block is about to move, called the impending motion, be- cause the current applied force is the largest force resistable by the physically largest generated friction force Fmax , governed by eq.
From the equation, the maximum friction force depends on the normal force and the static coeffi- cient of friction. The implication of the impending motion is the validity of the equilibrium condition and the friction force reaches the maximum value.
If the applied force is further increased, the block is no longer in the equilib- rium. That is the frictional force is now not enough to hold the block at rest. From the unbalanced resulting force, the block will then start moving in the di- rection of the applied force. Moreover, the magnitude of the friction force itself is known to be dropped off from the maximum value Fmax. Practically, the friction force decreases as the relative velocity of the mating surfaces increases.
In summary, there are three regions of the friction force development in tran- sitioning of the object from rest to motion. These are 1. No Motion is the region up to the point of slippage or impending motion.
Friction force is determined by the equations of equilibrium because the system is in equilibrium. Impending Motion is the moment where the body is on the verge of slipping.
Static friction force reaches the maximum value. Motion The body starts moving in the direction of the applied force. It will drop further with higher velocity. The smaller the value, the smoother the surfaces and the easier to move relatively to each other. Furthermore, this coefficient has another interpretation according to eq. When the friction force reaches the maximum value, i.
In other words, the direction of the reaction force is known in case of the impending motion. This is helpful in solving the problem as will be illustrated in some examples. Note the friction force is independent of the apparent or projected area of contact.
It involves the identification of the status of the motion that influences the determination of the friction force. Condition of impending motion is known to exist The body is in equilibrium and on the verge of slipping.
Unknown status of the problem This is the difficult case. The prob- lem is tackled by first assuming the system is in static equilibrium and using the equilibrium condition to determine the required friction force F. The result is then investigated to conclude the validity of the equilibrium assumption: 1. The assumed static equilibrium condition still holds and so the motion impends. Consequently, the equilibrium assumption is invalid and motion occurs instead.
And the equilibrium conditions are no longer held. The motion is accelerated. Using the above guidelines, the following representative problems are now presented and solved. Example 4. Solution: From the given statement, the block is on the verge of slipping. According to the free body diagram in fig. From the right free body diagram shown in fig. The coefficient of static friction for the contact surface is 0. Solution: From the setup configuration, if the counterweight is increased too much, the block will start sliding upward.
On the contrary, if the coun- terweight is decreased too much, the block will start sliding downward, lifting the balance upward. The problem asks the maximum and the minimum weight which still neither makes the block sliding upward or downward. This implies the system in question is on impending. The situations will be splitted into two cases; the block starts moving upward and downward, respectively. In both cases, however, forces in the normal direction are the same.
The normal force can be determined first by recalling the equilibrium condition in the normal y-direction. From fig. Block starts moving upward. The resist- ing friction force pointing downward fig. Block starts moving downward. The re- sisting friction force pointing upward fig.
The coefficient of static friction is 0. The force are applied with the block initially at rest. Solution: After reading the problem statement, the motion status of the block cannot be deduced. Additionally, it is not known whether the block is likely to move upward or downward the incline. Since this information is not known, it will be assumed as well.
Therefore the friction direction is assumed downward. After drawing the free body diagram as depicted in fig. Therefore the friction force is Therefore the friction direction is assumed upward.
After drawing the free body diagram as shown in fig. Therefore this required friction force cannot be fulfilled. The equi- librium assumption is thus invalid and the block is moving downward instead. Solution: In the first question, as the point of application of the applied force P increases, the block will be more likely about to tip over. When the block is about to tip over, the resultant supporting force from the ground must be acting through the last contact point at the far corner of the block, labeled as point A in fig.
For this problem, the block is being moved with a constant velocity. Therefore the developed friction is the kinetic friction force. Since the block is in equilibrium moving with constant velocity , all three forces acting on the block must meet at the common point denoted as B in the figure.
The location of this point will dictate the height h of the applied force P. Nevertheless, the system is still in equilibrium and hence three forces acting on the block must also meet at the common point. In this case the point is known; it is the center of gravity point G. The upper block is prevented from moving by a wire which attaches it to the fixed support.
The coefficient of static friction for each of the three pairs of mating surfaces is shown. Determine the maximum value which P may have before any slipping takes place. Solution: From the arrangement of the blocks, the 30 kg block cannot be moved. Only either the 50 or the 40 kg block, or together both blocks can be moved. However the applied force P acts at the 50 kg block, it is unlikely that the 40 kg block will move alone. Therefore there will be only 2 possible cases: either the 50 kg block alone or the 50 and 40 kg blocks will move together as one unit.
After drawing the free body diagram shown in fig. The block will not slip had either one alone reaches the maximum value. Consider the 40 kg-block.
Therefore the 40 kg-block cannot stay still and the requirement of no slippage cannot be satisfied. Hence this case will not happen. Therefore the 50 and 40 kg-block will not slip relative to each other and be at rest. To determine the maximum value of the applied force P , apply the equilibrium condition at the middle block. If the coefficient of static friction is 0. Also find the maximum value of x for which the bar will not slip.
Solution: The most distinct aspect of the system is that the light bar is a two-force member. The alternative view to this constraint is that the direction of the reaction force must reside in the static friction cone. Direction of the reaction force is along the propped bar. That reaction force is at the upper end where the coefficient of friction is lower. The lower reaction force will still be inside the static friction cone.
Therefore the developed friction force must reach the maximum value. For both cases, the free body diagrams in fig. Because the motion impends, the equilibrium equation of each block along the tangential direction can be set. The applied force P and the cable tension T can then be determined.
If both blocks are about to go together, the cable will get slack. Using the equilibrium conditions along the tangential direction on each block, the applied force P and the cable tension T can be determined. Solution: The status of the bar is impending motion. Solution: After careful consideration of the sketch in fig. The top most roller will then start falling down.
This described motion is shown in fig. The slipping will occur if one or more contacts of the lower rollers impend to slip. To understand the phenomenon, the free body diagram of the lower left roller is drawn as depicted in fig. Consequently, FA will reach the limit value before FB. FB , which may be less than FBmax , will be determined by the equilibrium equation.
The roller is recognized as a three-force member. Therefore the line of action of these three forces must intersect at a unique point for the system to be in equilibrium.
The only possible meeting point is B. From the geometry of the triangle OAB sketched in fig. If the coefficients of static and kinetic friction between the roll and the vertical barrier of the truck and between the roll and the incline are both 0.
Solution: The moving paper roll is making contacts with the vertical bar- rier of the truck and the incline. This problem is difficult because it is not clear from the statement how the roll is moving.
After drawing the free body diagram of the roll shown in fig. A and B both slip 2. Only B slips 3. Only A slips Each case has been investigated with the assumption of dynamical equilib- rium. After the calculation, only the last case — i. By observation of the equilibrium truck, the required tractive force P must balance the normal force NA.
When forces are applied over a region whose dimensions are not negligible com- pared with other pertinent dimensions, we must account for the actual manner in which the force is distributed by summing up the effects of the distributed force over the region.
For this purpose, we need to know the intensity of the force at any location and we will use the integration to determine their total effect. Figure 5. Principles of Lifting Machines Simple Lifting Machine Support Reactions Analysis of Perfect Frames Analytical Method Analysis of Perfect Frames Graphical Method Equilibrium of strings Virtual Work Linear Motion Motion Under Variable Acceleration Relative Velocity Projectiles Motion of Rotation Combined Motion of Rotation and Translation Simple Harmonic Motion Laws of Motion Motion of Connected Bodies Helical Springs and Pendulums Collision of Elastic Bodies Motion Along a Circular Path Balancing of Rotating Masses
0コメント