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Concurrent
and Parallel
Forces (^) , -.. ...-
force (FORS) II.: a physical quantity IIwl can affect the motion of an object.
OBjECTIVES
. Identify forces as vectors. . Define and calculate resultant and equilibrant forces, . Resolve forces into components. . Define and identify frictional forces_ . Solve problems involving frictional forces. . Define and calculate torque. . Solve motion problems by applying the two conditions of equilibrium.
A force never exists by itself. 70
COMPOSITION OF FOnCES
4.1 Describing Forces In Chapter 3, \ve studied the re- lationship between forces and motion. Now we shall take a look at some other characteristics of forces, and see how these characteristics are used in solving force problems. When you push a door shut with your hand, your hand exerts a force on the door. The door also exerts a force on your hand. When you sit in a chair, you push on the chair and the chair pushes on you. In both of these cases, there is physical contact between the objects that are exerting forces on each other. Forces can also be exerted without such physical con- tact. While an object is falling toward the earth, the earth exerts a gravitational force on the object and the object exerts a gravitational force on the earth. Yet there is no physical contact between the earth and the falling object. These examples illustrate several important characteris- tics of forces:
1. A netforce will change the state of lIIotion of all ollject. The
door moves because the force exerted by your hand is suf- ficient to overcome friction and other forces acting on the door. An object falls because a force is pulJing it toward the earth. As we saw in Chapter 3, the application of a net force to an object always produces an acceleration.
2. Forces can be exerted through long distances. Gravita~
tiona1 and magnetic forces have this characteristic.
3. Forcesalways occur in pairs. When one object pushes
f CONCURRENT AND PARALLEL FORCES
or pulls on another object, there is a force on each of the
two objects. In the given examples, the two objects were your hand and the door, you and the chair, and the falling object and the earth.
4. In each pair of forces, the two forces act in exactly opposite
directions. You pushed on the door, and the door pushed back. You pushed down, and the chair pushed up. The earth pulled the falling object toward the earth's center, and the object pulled the earth toward the object's center. Now let us see how the magnitudes of forces are mea- sured. (You will remember from Chapter 3 that forces are vector quantities and have both magnitude and direc- tion.) When an object is suspended from a spring, it is pulled toward the earth by the force of gravitation. The spring stretches until the restoring force of the spring is equal to the force of gravitation on the object. Another object having the same weight stretches the spring by the same amount. Both objects together stretch the spring twice as far, and an object with three times the weight stretches it three times as far, etc. This characteristic of coiled springs, that the amount of stretch is proportional to the force pulling on the spring, means that we can use the amount of stretch to measure the size of a force. A
device that measures forces in this way is called a spring
balance. The results of such measurements can always be expressed in terms of newtons. f r ; f f r f I f f I
4.2 Combining Force Vectors Since forces have magni-
tude and direction, and can combine like displacements, they are vector quantities. When a vector is used to repre- sent a force, the magnitude of the force is represented by the length of the arrow. The direction of the force can be deduced from the physical situation. For example, sup- pose a barge is being towed through still water by a tug- boat. The tugboat applies a force of 10 000 N to the barge through the towline. The length of the arrow representing the force is proportional to the magnitude of the force, 10 000 N. Figure 4~2 is a diagram of this example. The point of application of the force is the point at which the rope is attached to the barge. A long rope can transmit only a pull in a direction along its length. It cannot trans- mit a push or a sideways force. The rope is in the direction of the force. The arrow shows this direction. Force vectors are treated like velocity vectors. For exam- ple, suppose two tugboats are attached to the same barge. Tugboat A is pulling with a force of 10 000 N and tugboat B is pulling with a force of 7500 N in the same direction.
Figure 4-1. This telescoping boom crane can lift weights of more than 17 000 newtons to a height of sixteen stories. Refer to Section 1.8 for a discus- sion of spring and platform bal~ at/ces. 10000N Figure 4-2. A vector diagram of a force of 10 000 N applied by a tugboat to a barge through a tow~ line. The direction and point of application of the vector represent the line of action and point of application of the force.
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~. CONCURRENT AND PARALLEL FORCES sides of a parallelogram arc equal and that the diagonal of a parallelogram divides it into two congruent triangles. Vectors at right angles are a special case, as shown in the following example.
1':Xi:"'I.I~ Calculate the (a) direction and (b) magnitude of the resul- tant of the two forces acting at right angles on point 0 in Figure 4-5.
Solution
(a) Working equation: (J (^) = arctan (;:) = arctan = 33.7° east of south F (b) Workill'< equation: Fj{ = --L = '- cas () 15.0 N
~ 18.0 N
o lO.ON FE (
1O.0N
)
15.0 N.
Z ~IF,,>
I, (^) The angle between two forces acting on the same point is often not a right angle (see Figure 4-6). The parallelo- gram is completed as shown. Observe that the resultant is
drawn from the point on which the two original forces are acting
since the resultant will also act on this point. The magnitude of
FR is found graphically to be 23 N. The direction is 3CP south of cast. (Compare these values to the corresponding ones in the example above.) The resultant is very different if thc angle betwecn thc same two forces is 140.0°. The paral1elogram is constructed to scale in the same manner. The force vectors are the sides and the angle between them 140.0°. This parallelo- gram is shown in Figure 4-7. The diagonal must be drawn from 0, the point at which thc two forces act. The graphic solution for FR yields 10 N 8° west of south. The resultant of two forces acting at an acute or obtuse angle can be found trigonometrically by the laws of sines and cosines. The use of these equations is shown in the following example. o , , , , , , , , , , , , Figure 4-6. The resultant when two vectors are at acute angles. o (^) ~ , /^ / / / fFR (^) ///
/^ / Figure 4.7. The resultant when two vectors are at obtuse angles.
74 CHAPTER^4
EX.\I\1PLI:: Calculate the (a) magnitude and (b) direction of the resul- tant of the two forces acting at an angle of 50.0" on point 0, as shown in Figure 4-6.
~!!!!.!j'!"~~::!r!!!,"~1i"!J""""~ (^) _ L Given^ .__~nknown
FE = 10.0 N FI{
FA ~^ 15.0^ N^ e
e' ~^ 130.0'
..8 811_.. II.!. Basic equations I ~ r-2: 2 ~-~- , FR = vFE + FA - 2FEFA COS f)
FA FR --~- sin f}^ sin (J'
Solution
(a) Working equation:
FR = VFE~ + TA2 - 2FEFA cos fJ'
~ /(10.0 N)' + (15.0 N)' - 2(10.0 N)(15.0 N)(cos 130.0') ~ 22.8 N
(b) Working equation: sin f) =
FA sin ()'
Fe
. [
(15.0 N)(sin 130.0') H=arcsm (^) (22.8 N) ]
= 30.3Q^ south^ of east
I)RACI'IU~ pnOBLEllS 1. Two forces act concurrently at right angles
on point O. One force of 30.0 N acts south. The other force acts west with a magnitude of 40.0 N. Calculate the magnitude and direction of the resultant. Ans. 50.0 N 36.90 south of west
- A force of 3.50 N acts north on point O. A second force of 8.75 N acts concurrently on point 0, but at an angle of 30.00 west of north. Calculate the magnitude and direction of the resultant force. Ails. 11.9 N 21.60 west of north
An equilibrant force is equal in magnitude to the resultant of two or more concurrent forces and
acts in the opposite direction.
4.3 The Equilibrant Force Equilibrium is the state of a
body in whichthere is nochangein its motioll. A body in equi-
librium is either at rest with respect to other bodies or moving at constant speed in a straight line. In this section we shall discuss the conditions for equilibrium of bodies at rest. The same conditions hold for the equilibrium of bod- ies that are in motion.
- Two soccer players kick a ball at the
same instant. One strikes with a force of 65 N north and the other 88 N east. Find the resultant force on the balL
- Find the magnitude of the resultant of
the two force vectors shown in Figure 4-7 by the trigonometric method. G)Two ,- children^ pull^ a wagon^ by exerting
forces of 15 Nand 18 N at the same
point. If the angle between them is 35.0°, what is the magnitude of the resultant force on the wagon?
- A boy and a girl carry a 12.0-kg bucket
o Figure4.9. Adiagram for the de- termination of the perpendicular components of a force. A single force can be resolved into two or more components that have the same effect. o A (^) I, FWD
Figure 4-10. Resolution of gravita- tional force. One component acts parailel to the plane while the other component acts normal (per- pendicular) to the plane. .- CHAPTER 4 of water by holding the ends of a rope with the bucket attached at the mid- dle. If there is an angle of 100.00 be- tween the two segments of the rope, what is the tension in each part?
- Three men are pulling on ropes at-
tached to a tree. The first man exerts a force of 6.0 N north, the second a force of 35 N east, and the third 40 N 30.0° east of south. (a) Find the resul- tan t force on the tree by using the graphic solution. (b) What is the equi- librant force? RESOLUTION OF FORCES
4.4 Components of Force Vectors Frequently a force
acts on a body in a direction in which the body cannot move. For example, gravitational force pulls vertically downward on a wagon on an incline, but the wagon can move only along the incline. Finding the magnitude of the force that is pulling the wagon along the incline is an ex- ample of resolution of forces. Instead of a single force, two forces acting together can have the same effect as the origi~ nal single force. One of the forces can be parallel to the surface of the incline and can pull the wagon along the incline. The other can be perpendicular to the surface of the incline. This force does not contribute to the force along the incline. These forces are at right angles to each other. As we saw in Section 2.11, two vectors that have the same effect as a single vector are called the components of the original vector. This procedure of finding component forces is called resolution of forces. Most of the examples we shall consider involve resolving a force into components that are at right angles to each other. Before working problems with objects on an incline,
study the following example, in which a force is resolved
into two perpendicular components. , j 4.5 Resolving Gravitational Forces An object placed ~ on an inclined- plane is attracted by the earth. The force of c attraction is the weight of the object. See Figure 4-10. The plane prevents the motion of the object in the direction of Fw, the direction in which the earth's attraction acts. The vector representing the force of attraction can, however,
be resolved into two components. One component acts in
a direction perpendicular to the surface of the plane. In physics, the term normal is often used to mean perpendic- ular. Hence we label the normal component FN' The^ other
, f CONCURRENT AND PARALLEL FORCES component, Fp, acts parallel to the plane. We choose these two components because they have physical significance. The vector FN represents the force exerted by the object perpendicular to the incline or the amount of the object's weight supported by the incline. The vector Fp represents the component that tends to move the object down the incline. (The plane is assumed to be frictionless.) Using Fw as the diagonal, we can construct the parallel- ogram ODEF and find the relative values of the sides Fp and FN by plotting to scale. We can also express these val- ues trigonometricaJly. Since right triangles ABC and OED have mutually perpendicular, or parallel, sides, the trian- gles are similar and LEOD=O. Hence sin 0 can be ex- BC Fp pressed either as - or -. This equation means that AB Fw
the force vector parallel to the plane, FI" is smaller in mag-
nitude than the weight vector, Fw, in the same ratio as the height of the plane, BC, is smaller than its length, AB. By using cos 0, it may be similarly shown that the mag- nitude of the force vector perpendicular to the plane, FN is related to the weight vector of the object in the same \vay that the base of the plane is related to its length. If 0 and Fw are known, Fp = Fwsin 0 FN = Fwcos 0 f I r, ,,
~
and
Making the plane steeper increases the component Fp and decreases the component (^) FN' This steeper inclined plane is shown in Figure 4-11. The vector FA, which is equal and opposite to Fp, represents the applied force needed to keep the object from sliding down the plane. The steeper the plane, the greater this force becomes. It should be noted that in both Figure 4-10 and Figure 4-11, the normal force that the plane exerts on the object in reac- tion to force FN is not shown, since it is not involved in the calculations for finding FA, 77 B A E c Figure 4-11. When the angle of the incline increases. the compo- nent of the weight acting parallel to the plane increases. while the component that acts normal (per- pendicular) to the plane decreases. EX;\MPU~ A force of 10.0 N acts on point 0 at an angle of 37.0° east of south, as shown in figure 4-9. find the magnitudes of the (a) eastward and (b) southward components of its force vectors. I Given h ~^ 10.0 N o = 37.0Q Unknown FE Fs Basic eqnation", (^) ~ sin ()~^ FE/FR 'cos 8 FslFR
GROUP B
- Two paramedics are carrying a person 7. A traffic light is supported by two
on a stretcher. One of the paramedics wires, as shown in Figure 4-13. If the exerts a force of 350 N at 58° above the maximum tension in each wire is horizontal and the other exerts a force 750 N, what is the maximum weight of of 410 N at 4Y above the horizontaL the light they can support? What is the total upward force exerted @ 2.00^ X^ 1OJ_kg car is to be held^ on^ a by the paramedics? 20.0° incline by a rope in which the
- Ms. Jones has attached a sign that has maximum tension is 8.00 x 103 N.
a weight of 495 N to a wall outside her (a) Will the rope support the car?
office, as shown in Figure 4-12. Deter- (b) If the rope is released, how far will mine (a) the magnitude of the tension the car have moved down the incline in the chain and (b) the thrust force by the time its speed reaches 35.0 m/s? exerted by the rod. , CONCURRENT AND PARALLEL FORCES , , f r f f I r l l Figure 4-12. FRICTION 4.6 The Nature of Friction In Section 4.5, we dis- cussed the resolution of the weight of an object resting on an incline. One of the components of the weight tends to pull the object down the incline. As the angle of the incline increases, this component also increases. The slightest angle of incline will produce the component that pulls the object down the incline if there is no restraining force on the object. However, in performing experiments of this type, we find that the object does not begin to slide until the component parallel to the incline reaches a certain value. This means that forces must exist between the ob- ject and the incline that prevent the object from sliding.
These forces are called forces of friction, or simply friction.
Friction is a force that resists motion. It involves objects that are in contact with each other. The causes of friction are sometimes complicated. Take, for example, a book lying on a table: the book's weight slightly deforms the surface of the table, along with that of the book. A "plowing" force is required to move the book
o
o o
Figure 4-13. \
Without friction you couldn't write your homework.
Fw'
Fw
Figure 4-14. The force on a block being pulled along a surface Ff is the force of sliding friction. The block does not move until FA ex- ceeds the force of starting fric- tion, which is usually greater than F,.
CHAPTER 4 over these deformations. The irregularities on both sur- faces tend to interlock and offer resistance to the sliding of the book. In the process, tiny particles are torn from one surface and become imbedded in the other. From this example one would expect that if the two sur- faces are carefully polished, sliding friction between them wou1d be lessened. Experiments have shown, however, that there is a limit to the amount by which friction may be reduced by poJishing the surfaces. 1£they are made very smooth, the friction between them actually increases. This observation shows that some cases of sliding friction are caused by the forces of attraction between the molecules of substances. In many instances, friction is very desirable. We would be unable to walk if there were no friction between the sales of our shoes and the ground. There must be friction between the tires of an automobile and the road before the automobile can move. When we apply the brakes on the automobile, the friction between the brake linings and the brake drums, or disks, slows down the wheels. Fric- tion between the tires and the road brings the car to a stop. In a less obvious way, friction holds screws and nails in place and it keeps dishes from sliding off a table if the table is not perfectly level. On the other hand, friction can also be a disadvantage, as it is when we try to move a heavy piece of furniture by sliding it across the floor.
4.7 Measuring Friction Friction experiments are not difficult to perform, but the results are not always easy to express as equations or laws. The following statements, therefore, should be understood as approximate descrip- tions only. Furthermore, they deal exclusively with solid objects. Frictional forces involving liquids and gases are beyond the scope of this book. Also, our discussion is re- stricted to starting and sliding friction. Starting friction is the maximum frictional force between stationary objects. Sliding friction is the frictional force between objects that are sliding with respect to one another. Static friction (which varies from a to the value of starting friction) and rolling friction are not considered in this text.
1. Friction acts parallelto the surfaces that are in contact and
in the direction opposite to the motion of the object or to the net force tending to produce such motion. Figure 4-14 illustrates this principle. The weight of the block, Fw, is balanced by the upward force of the table, Fw'. The force FA is sliding the block along the table top. In this case, FA is parallel to the table top. The sliding frictional force, Ffl also parallel
Figure4-16. Friction between skis and snow is appreciably reduced with the application of a wax layer on the wood or metal surface of the skis.
The squeaking wheel gets the grease.
CHAPTER 4 mal (perpendicular) force between the surfaces. (Within certain limits, this equation is an approximate summary of friction measurements.) The coefficient of starting friction is determined in a similar fashion, except that Ff is the
force of starting friction. The approximate values for the
coefficients of friction of various surfaces in contact with each other are given in Table 4-1. 4.8 Changing Friction In winter, we sand icy sidewalks and streets in order to increase friction. Tire chains and snow tires are used fOf the same reason, In baseball, pitch- ers often use rosin to get more friction between their fin- gers and the ball. Many more examples could be given in which friction is purposely increased by changing the na- ture of the surfaces that are in contact. The most common method of reducing sliding friction is by lubrication. The skier pictured in Figure 4-16 applied a layer of wax to the skis to reduce friction. A thin film of oil between rubbing surfaces reduces friction. The lesser fric- tion between a liquid and a solid has replaced the greater friction between two solids. Alloys have also been devel- oped that are in effect self-lubricating. For example, when steel slides over an alloy of lead and antimony, the coeffi- cient of friction is less than when steel slides over steel. Bearings lined with such an alloy reduce friction. From Table 4-1 it is also obvious that if a bearing is coated with a plastic such as Teflon, there is very little friction. Such bearings are used in electric motors where the use of a liquid lubricant is undesirable. Friction may also be greatly reduced through the use of ball bearings or roller bearings. Sliding friction is changed to rolling friction, which has a much lower coefficient. Using steel cylinders to roll a heavy box along the floor is another example of changing sliding to rolling friction.
4.9 Solving Friction Problems The force required to slide an object along a level surface can be computed easily from the weight of the object and the coefficient of sliding friction between the two surfaces. However, when the force applied to the object is not applied in the direction of the motion, it is necessary to resolve forces in the calcuJa- tion. This resolution of forces is illustrated in the following example. In the case of an object resting on an incline, the forces of starting and sliding friction will determine whether the object remains at rest, slides down the incline with con- stant speed, or accelerates as it descends. How the angle of the incline and coefficient of sliding friction are used in such a problem is illustrated in another example.
I
~ I I, ~" r ~ CONCURRENT AND PARALLEL FORCES^83 EXAMPLE A box weighing 450 N is pulled along a level floor at con- stant speed by a rope that makes an angle of 30.0~ with the floor, as shown in Figure 4-17. If the force on the rope is 260 N, (a) what is the horizontal component (Ph) of this force? (b) What is the normal force (FN)?
(c) What is the coefficient of sliding friction (J.L)?
~';~'~'I!' I :;:;-: 'Gi~en
. (^) -~__"IU ~
Unknown --- Fw = 450N e = 30. =260N . Basic equations F, cos8=-
FA
FN=Pw-FAsin
F,
/-'=- FN
Fh FN
Solution
(a) Working equation: Ph = FA ~os ()
= (260 N)(0.866)
= 225 N (b) The normal force (FN) is the difference between the downward force of the block's weight (Fw) and the vertical component of the force of the rope (Fv): Motion
. (^) -
Working equation: FN = Fw - FA sin ()^ F !A::= 260 N ~ 452 N - (260 N)(0.500) ~ i ~ 320 N (^) '
F 1 ',30.0 '
-~Fh Ff Fh (c) Working equation: (^) J.L = FN = FN 225N ~
320N
= 0.703 Fw=450N
I~X:;\WU~ A wooden block weighing 130 N rests on an inclined plane, as shown in Figure 4.18. The coefficient of sliding friction between the block and the plane is 0.620. Find the angle of the inclined plane at which the block will slide down the plane at constant speed once it has started moving.
f
N 8
, ,
r
I. , f CONCURRENT AND PARALLEL FORCES
PROBLEMS: GROUP A
- A horizontal force of 400.0 N is re-
quired to pull a 1760-N trunk across the floor at constant speed. Find the coefficient of sliding friction.
- How much force must be applicd to
push a lo3S-kg book across the desk at constant speed if the coefficient of sliding friction is 0.
- A force of 105 N is applied horizon-
tally to a 20.0-kg box to move it across a horizontal floor. If the box has an acceleration of 3.00 mJs2, find the coefficient of friction.
- A 150Q.Q-N force is exerted on a
200.0-kg crate to move it across the floor. If the coefficient of friction is 0.250, what is the crate's acceleration?
- A IOO.O-kg commuter is standing on a
train accelerating at 3.70 m/S2. What coefficient of friction must exist be- tween the commuter's feet and the floor to avoid sliding?
CROUP B
- A 146-N force is used to puJI a 350-N wood block at constant speed by a rope making an angle of 50.0° with the floor. Find the coefficient of slid- ing friction.
- A 75.0-kg baby carriage is pushed along a level sidewalk by exerting a
PARALLEL FORCES 85 force of 50.0 N on the handle, which makes an angle of 60.0° with the hori~ zontal. What is the coefficient of fric- tion between the carriage and the sidewalk?
- A 3.00-kg wood box slides from rest
down a 35.00 inclined plane. How long does it take the box to reach the bottom of the 4.75-m wood incline? (See Table 4-1 for the coefficient of friction.)
- A 65.0-kg crate is to be accelerated at
7.00 m/s2 up an incline making a 25. angle with the horizontaL If the coef- ficient of sliding friction between the crate and the incline is 0.200, how much force is required?
- A 60.0-kg crate is attached to a
weight by a cord that passes over a frictionless pulley, as shown in Figure 4.19. (a) If the coefficient of friction is 0.500, what weight will keep the crate moving up the 40.00 incline at a con- stant speed? (b) If the cord is cut when the crate is at rest at the top of the incline, how far would the crate have slid by the time its speed reached 7.50 m/s?
- If the coefficient of friction between a
set of waxed skis and the snow is D.10, at what angle will a 9D.D-kg skier move at a constant speed down the slope? 4.10 Center of Gravity Thus far in our study of forces, we have been treating all the forces acting on a body as if they were acting at a single point. However, there can be many forces, each acting at a different point on the object. For example, Figure 4-20 represents a stone lying on the ground. Since every part of the stone has mass, every part is attracted to the center of the earth. Because of the large size of the earth, all the downward forces exerted on the stone are virtually paralleL The weight of the stone can be thought of as a force vector that is the vector sum, or re- sultant, of all these parallel force vectors. Parallel forces
86
Figure 4-20. The stone's center of gravity is the point where all the weight seems to be concentrated.
F,
c
Fw Figure 4-21. The weight of the bar, Fw. is apparently concen- trated at the center of gravity, C, and can be balanced by an equal and opposite force, FE. applied at C.
B A
F.
Figure 4-22. Two parallel forces. FA and Fw. act at different points. A tendency to rotate results.
CHAPTER 4
act in the same or in opposite directions at different points on an object. The resultant of parallel forces has a magnitude
equal to the algebraic sum of an the forces. The resultant
acts in the direction of this net force. But where in the stone is this resultant force acting? Experiments show that if the proper point of application is chosen, the stone can be lifted without producing rota- tion. As shown in Figure 4~20f the equilibrant lifting force vector is then in line with the resultant (weight) vector of the stone. In other words, the stone acts as if all its weight
were located at one point, which is called the center ofgrav-
ity. The center of gravity of any object is that point (It which all
of its weight can be considered to be concelltrated. In Figure 4-21, the center of gravity of the bar is at C. In a bar of uniform construction, C is at the geometric center. But if the density or shape of the bar is not uniform, C is not at the geometric center. Since the weight of the bar, F," can be considered to be acting at C, the bar can be suspended without changing its
rotation by an equilibrant force, FE, applied at C. Since Fw
and FE are equal but opposite vectors, they counterbalance each other. In this condition, the bar is in both translatiOlwl and rotational equilibrium. This means that the bar is not accelerating and its rotation (if any) is constant.
4.11 Torques The two forces represented in Figure 4- by the vectors FA and FR are parallel. They do not act on the same point as did the concurrent forces we studied earlier in this chapter. To measure the rotating effect, or torque, of such parallel forces in a given plane, it is first necessary to choose a stationary reference point for the measurements. We shall refer to this stationary reference point as the pivot point. ' Sometimes, as in the case of a seesaw, there is a "natu- ral" point about which the rotating effects can be mea- sured. However, such a pivot point is "natural" only when the seesaw is in motion. When it is motionless there is no "natural" pivot point. Any point on the seesaw, or even bevond it, can be chosen. Once -'a suitable^ pivot^ point^ is chosen,^ a perpendicular line is drawn on the vector diagram from it to each of the lines along which force vectors act on the object. Each
such line is called a torque arm. In some cases, the force
vectors must be extended in order to meet the perpendicu-
lar. Torque, T, is the product of (I force and the length of its
torque arm. The unit of torque is the meter-newton.
To illustrate the concept of torque, consider the bar in Figure 4-22 with the application of an additional force, Fc,
Torque (^) ,arm
A'~ ~-------
I I I I I I A
Figure 4-25. The torque arm is the perpendicular distance from the pivot point to the line indicat- ing the direction of the applied force.
FN (^) ~FA I I L ..J
Figure 4-26. The resolution of forces is used to find the torque produced by a force acting on the bar at an angle other than perpen- dicular.
'F' Figure4-27. A pair of parallel forces of equal magnitude acting in opposite directions but not on the same point is called a couple. (The weight of the bar is not con- sidered in this example.)
F
CHAPTER 4
against2iJ N + 10 N downward). The sum ofthe clockvdse torques is equal to the sum of the counterclockwise
torques. Two methods of computing the torques are
shown. In the upper drawing of Figure 4-24, the left end of the bar is chosen as a pivot point. In the lower drawing, the point of application of the upward force is used as the pivot point. This simplifies the calculation. When forces are applied to a bar at an angle other than perpendicular, the distances along the bar measured from the points of application of the forces cannot be used to
measure the torque arms. The torque arms must always be
measured perpendicular to the direc60ns of the forces. Figure 4~25 shows such a situation. A meter stick is under the influence of three parallel forces. To find the torque arms, it is necessary to draw a horizontal line through the pivot point. Since tne weights hang vertically, a horizontal hne is perpendicular to the force vectors of the weights. The problem is simplified by placing the pivot point at the cen- ter of gravity. If the bar is not homogeneous, this center of gravity may not be located at tne geometric center of the bar. (Experimentally, the center of gravity can be approxi. mately located by finding the point where the bar bal- ances.) Placing the pivot point at the center of gravity eliminates two torques in the equation. It eliminates the torque produced by the weight of the bar located at some distance from the geometric center of the bar. And it also eliminates the torque produced by the force acting up-vard at C. The required torque arms, CA' and CB', are then found by multiplying the distances CA and CB by the co- sine of the angle ACA' or BCB'. In Figure 4-26, FA is applied to the right end of the bar at an angle other than perpendicular. In order to find the counterclockwise torque produced by (^) FA' we^ choose^ the left end of the bar as the pivot point. Then we find the vertical component, (^) FN' of^ FA,^ By trigonometry, FN=FA sin 0 Since FN is perpendicular to the bar, the required torque is T~FNI
Substituting, T=FAl sin 0 lienee, 1 sin 0 is the torque arm of FA, The result is further verified in Figure 4-26, where 1 sin 0 is the length of the perpendicular from the pivot point to tne extended line of direction of FA, This is in accord with the definition of torque arm in Section 4.11.
CONCURRENT AND PARALLEL FORCES
4.13 Coupled Forces The conditions of equilibrium hold true no matter how many forces are involved. An
interesting example is one in which two forces of equal mag~
nitude actin opposite directionsin the same plane, but not on the
same point. Such a pair of forces is called a couple. A dia- gram of a couple is shown in Figure 4-27. The torque is equal to the product of one of the forces and the perpen- dicular distance between them. (This can be proved by computing the sum of the torques produced by the action of the separate forces about any desired pivot point.) A good example of a couple is the pair of forces acting on the opposite poles of a compass needle when the needle is not pointing north and south. A couple cannot be balanced by a single force since this single force would be unbalanced and would produce lin- ear motion where it was applied. The only way to balance a couple is with another couple; the torques of the two couples must have equal magnitudes but opposite directions.
E c D -. I~I
SOON 10.00m
1.000N
EXiOWLE A horizontal rod, AB, is 10.00 m long. It weighs 500 Nand its center of gravity, C, is 3.00 m from A. At A a force of 1000 N acts downward. At B a force of 750 N acts downward. At D, 2.00 m from B, a force of 400 N acts upward. At E, 1.00 ill from A, a force of 750 N acts upward. (a) What is the magnitude and direction of the force that must be used to produce equilibrium? (b) Where must it be applied? (See Fig ure 4-28.)
Solution
(a) Consider the known upward force vectors as being positive and the known downward force vectors as being negative. The algebraic sum of these force vectors will then be the resultant force vector. This resul- tant force vector must be counterbalanced by a force vector in the oppo- site direction in order to establish translational equilibrium.
75DN + 400 N - 1000 N - SOD N - 750 N ~^ -1100^ N
Therefore, 1100 N must be applied upward to establish translational equilibrium. (b) Use A as the pivot point, and let x be the distance from A to the point where the 11OU-Nforce must be applied to prevent rotary motion. clockwise torgue ~^ (SOD^ N)(3.00^ m) + (750 N)(lO.OO m) counterclockwise torgue ~^ (750 N)(1.00^ m) + (40D N)(S.OO m) + (1100 N)x
9000 m.N ~^ 3950m.N + (1100 N)x
x = 4.59 m 4.59 m is the distance from A to the point where the 1100-N upward force must be applied.
