Download Forces and Equilibrium and more Study notes Physics in PDF only on Docsity!
Problem 4.1. In Fig. 4-l(a) two forces are shown acting at a point in an object. Find the magnitude
and direction of the single force that can replace those two forces and have the exact same effect.
Solution
In Fig. 4-l(b) the resultant Rand the replaced forces F 1 and F 2 (in dashed form), as well as F 2 shifted parallel to itself so that it is tail to head with F 1 (see Sec. 3.1 ). Since the two original forces are at right angles to each other, we can use the pythagorean theorem to obtain the magnitude of the resultant
force: R 2 = F 12 + F/ = (30 lb) 2 + (40 lb)2 = 2500 lb2. Taking the square root, we obtain R = 50 lb.
To get the direction of R we determine its angle () with the horizontal. We have tan () = opposite/
adjacent = 40/30 = 1.33 or () = 53 °. Thus R has magnitude 50 lb and acts at an angle 53 ° above the
horizontal.
The Resultant of a System of Forces
The vector sum of the forces acting on an object is called the resultant force on the object. The
laws of nature are such that when two or more forces are acting at the same point in an object, they can
be replaced by their resultant acting at the same point, which will have the same exact effect on the
object as the original set of forces.
Types of Forces
A force can be either due to direct contact (contact force) such as a hand pushing a block or a
rope dragging a box or due to influence from afar (action at a distance) such as the gravitational pull
of the earth on a satellite or the push of one magnet on another not in contact with it. On the human
scale there are many different forces of either type. But on the atomic scale there are only four
fundamental forces: gravitational, electromagnetic, weak nuclear, and strong nuclear-all of them
actions at a distance.
A force is a mechanical effect of the environment on an object. It is either a push or a pull on an
object, and has both a magnitude (in appropriate units such as newtons, dynes, or pounds-units of
force are discussed in detail in Chap. 5) and a direction. It can thus be represented by a vector. A force
has two basic effects on an object. (1) It can change the motion of the object, which is the subject of
Newton's famous second law (Chap. 5). (2) It can distort the shape of an object such as by stretching,
compressing, or twisting the object.
4.1 FORCES
Note. In introductory mechanics it is usually assumed that all forces act in the same plane
(usually called the xy plane). Such forces are said to be coplanar. This assumption
simplifies the mathematics considerably but still allows for a substantial understanding of
the underlying physics involved.
Forces and Equilibrium
Chapter 4
(b) Fig. 4-2 (a)
.. =30lb^1 F---^ B •······-^ D
but 4-l(a), we have the same two forces acting on a rigid body as in Fig. 4-2(a) In Fig. Problem 4.2.
and C. Can one still replace these two forces by a single B now they are acting at different points
resultant force that has exactly the same effect on the motion of the rigid body and, if so, give an
example of such a resultant force?
Line of Action
When a force acts at a point in an object, one can draw an imaginary line through that point and
parallel to the force. This is called the line of action of the force.
refers to an object that doesn't change its shape when forces act on it. No real object A rigid body
is truly rigid, but the concept is a good approximation for stiff objects. In studying the relation of force
and motion we will usually assume that we have rigid bodies. While in general the effect of a force on
a rigid body depends on where it acts, a force acting on a rigid body can be applied anywhere along its
line of action and still have exactly the same effect.
(b)
Fig. 4-
I
I
I ( 40lb
I
I
I
I
:..+J
I^ •
I
I
[CHAP. 4 FORCES AND EQUILIBRIUM
(a)
No. The translational motion of the eraser is a parabolic arc and that of the moon is a circle, whereas
approximate of example An in a straight line. must be for translational equilibrium the motion
translational equilibrium would be a block sliding on an ice-covered lake; the block would move in a
straight line without slowing down.
Does the motion of the eraser in Fig. 4-3 or of the moon in Fig. 4-4 correspond to Problem 4.6.
translational equilibrium?
Solution
Fig. 4-
I
I
I
I
Orbit
I
I
I
I
I
I
I
I
I
uniform Translational equilibrium means that the object as a whole, aside from rotation, has
translational motion, that is, its center of mass is either at rest or moving at constant speed in a straight
line.
The circular dashed line represents the translational motion of the moon. This moon has no
rotational motion since its orientation does not change. The moon, in effect, stays parallel to itself
throughout the motion.
Describe the translational and rotational motion of the cratered moon around the Problem 4.5.
planet in Fig. 4-4.
Solution
[CHAP. 4 FORCES AND EQUILIBRIUM^90
Problem 4.8. A uniform rod is connected to two cords that exert the only forces on the rod, as
depicted in Fig. 4-6; (i.e., we assume there is no pull of gravity on the rod). For each case determine
whether the rod is in translational equilibrium. If so, can it also be in rotational equilibrium?
Note. It is also possible to have rotational equilibrium without translational equilibrium, a
situation that will be discussed in a later chapter.
Equilibrium with Only Two Forces Acting
If the two forces F 1 and F 2 (see Fig. 4-5) are equal in magnitude and opposite in direction (that is,
F 1 + F 2 = 0), then the object is in translational equilibrium. If in addition the two forces act along a
common line of action (collinear forces), as in Fig. 4-S(b), then the object is also in rotational
equilibrium.
A totally isolated object (no forces) is in both translational and rotational equilibrium in an inertial
reference frame. However, even rigid bodies that do have forces acting on them can be in either
translational or rotational equilibrium, or both, under suitable conditions. The condition for
translational equilibrium is the statement of Newton :S, first law, also known as the law of equilibrium.
We give here some simple cases.
4.3 NEWTON'S FIRST LAW
A Frame of Reference refers to the "framework" that defines the coordinate system in which one's
measurements and observations are made. If a coordinate system is fixed to the earth and another one
is fixed to a rotating merry-go-round, one is going to observe things differently in each. Each of these
coordinate systems is fixed in a different frame of reference.
An inertial frame ofreference, by definition, is a frame ofreferencein which a completely isolated
object (no forces) will appear to be in both translational and rotational equilibrium. For most purposes
the earth can be considered an inertial frame; that is only an approximation, however, because the
earth spins on its axis=although it is a very slow spin-once every 24 h. The importance of inertial
frames is that Newton's laws hold only in such frames, and most of the other laws of physics take on
simpler form when described in such frames. We will always assume that we are describing things in
an inertial frame of reference unless otherwise indicated.
If the eraser were tumblingat a uniform rate, it would indeed be in rotational equilibrium; that, in
fact, is a good approximation to what happens if air resistance is not an important factor. The moon is
certainlyin rotationalequilibrium, since we are shown that the moon does not rotate at all.
Problem 4.7. Does the motion of the eraser in Fig. 4-3 and of the moon in Fig. 4-4 correspond to
rotational equilibrium?
Solution
Rotational equilibrium means that the object-whether it is undergoing translational motion or
not-is either not spinning or it is spinning in a uniform fashion. For simple symmetric objects this
means spinning at a constant rate about a fixed direction.
CHAP. 4] FORCES AND EQUILIBRIUM 91
While this method of solving a vector equation seems more cumbersome than the geometric method, it can be applied to more general cases where the geometric approach is too difficult to use.
0.5F 1 + 0.5F 1 = lOON or
Using F 1 = Fi in they-component equation gives
F 1 sin30° +F2 sin30° - lOON = 0 or 0.5F1 + 0.5F2 = IOON
Similarly, the y-component equation gives
F1 cos 30 ° - F2 cos 30 ° + 0 = 0 or
Substituting into the x-component equation,
F3x = 0
F3y = -lOON
Fix= -F2 cos 30 °
F2y = F2 sin 30 °
Fix = F1 cos 30 ° F,y = F1 sin30°
From Fig. 4-7(c), we have
and
(b) We now solve the problem algebraically. Choose. the x axis along the rod and the y axis perpendicular to the rod at its center. Now slide the vectors parallel to themselves to the origin, for easier visualization Fig. 4-7(c). Since the vector sum of the three forces equals zero, we must have for the components
(b) (c) Fig. 4-
(a)
y
(a) Newton's first law tells us that the resultant of the three forces acting on the rod must be zero. In Fig. 4-7(a) we redraw the rod as an isolated object and include only the forces acting on it (body
diagram). The condition F 1 + Fi + F 3 = 0 implies that the thre 7 vectors, drawn head to tail, form a
closed triangle. As can be seen in Fig. 4-7(b), the triangle is equilateral for our case, so
F1 = Fi = F3 = I 00 N.
Solution
CHAP. 4) FORCES AND EQUILIBRIUM 93
Fig. 4- (b) (a)
elephant and a teenager are having a tug-of-war, as shown in Fig. 4-8(a). Does An Problem 4.12.
Newton's third law imply a draw?
Consider a book lying at rest on a horizontal table. Problem 4.11.
What are the forces on the book? (a)
What is the reaction force to each of these forces? (b)
What effect do the reaction forces have on the book? (c)
Solution
There are two forces acting on the book: its weight (the downward pull of gravity toward the center (a)
of the earth) and the force exerted upward on the book by the tabletop.
The reaction to the weight is an upward pull of equal magnitude exerted on the earth by the book. (b)
The reaction to the table's force is a downward push of equal magnitude on the table by the book.
The reaction forces have no effect on the book! By definition, any effect on the book is represented (c)
on the book. The reaction forces act on the earth and on the table-not book. the on by a force
exerts a A states that if some object law of action and reaction, This law, otherwise known as the
that is equal in magnitude and A on object Fba exerts a force B then object B, on object Fab force
The law holds both for contact forces and for action-at-a-distance -Fab· = Fba opposite in direction:
forces.
NEWTON'S THIRD LAW 4.
with Any Number of Forces Equilibrium
of forces, we again have two conditions for equilibriwn. The n For the general case of any nwnber
which says that the vector sum Newtons first law, first is the condition for translational equilibrium, or
- For small objects or particles, where rotation can be ignored, it is the = LFi of all the forces is zero:
only condition of equilibrium. For extended objects, the second condition, for rotational equilibrium,
is again needed. The general case of rotational equilibrium will be discussed in a later chapter. The
rest of this chapter is concerned only with translational equilibrium.
[CHAP: 4 FORCES AND EQUILIBRIUM 94
Static Friction
When two surfaces are at rest with respect to one another, the frictional force each exerts on the
other always opposes any tendency to relative motion. The frictional force on an object adjusts itself
in magnitude and direction to oppose and counterbalance any other forces on the object that would
varies, as needed, from zero magnitude up to some maximum It tend to make the object start to slide.
The maximum (fs)· value to stop such slippage. Such a frictional force is called a static friction force
Normal Force
force because it acts The. force responsible for this "pressing together" is called the normal
perpendicular to the two surfaces. By Newton's third law each surface exerts a normal force that is
equal in magnitude and opposite in direction to that exerted by the other. Figure 4-10 indicates the
frictional and normal force on each object when a block is in contact with an inclined plane. The
frictional force (parallel to the surface) and the normal force (perpendicular to the surface) acting on a
surface can always be thought of as the components of the overall force acting on that surface due to
the other surface in contact with it.
is the rubbing force between two objects whose surfaces are in contact. The force of Friction
friction always acts parallel to the touching surfaces. By Newton's third law each surface exerts a
frictional force that is equal in magnitude and opposite in direction to that exerted by the other. The
magnitude of the frictional force exerted by each surface on the other depends on how tightly the two
surfaces are pressed together.
FRICTION 4.
15 N hangs at the end of a (weightless) cord suspended from = w A block of weight Problem 4.13.
the ceiling. What is the tension in the cord, and with what force does the cord pull down on the
ceiling?
Solution
The tension is the same at all points of the cord and is equal to the magnitude of the force pulling at
either erid. Since the block is in equilibrium under the action of two vertical forces (the weight
downward and the pull of the cord upward), these two forces must have the same magnitude. Hence the
15 N. By Newton's third law the magnitude of the pull of the block downward upward pull of the cord=
also equals the magnitude of the pull of the T = 15 N. The tension w = T on the cord is also 15 N, so
ceiling on the cord, which by Newton's third law equals the pull of the cord downward on the ceiling.
Thus the downward pull of the top of the cord on the ceiling is the same as the downward pull of the
block on the bottom of the cord. Thus we see that a weightless rope transmits an applied force from one
end to the other.
There is, however, one circumstance where there is a common tension throughout the rope, and
this tension always equals the magnitude of the forces acting at the ends of the rope-whether the rope
is horizontal or vertical, whether the rope is in equilibrium or not. This is the circumstance where the
most problems one characterizes such a rope as a cord, string, or thread to In rope is weightless.
Obviously no cord is completely weightless, but if it is very light in indicate its "lightness."
comparison to the other objects in the problem, it can be assumed weightless without much error.
[CHAP. 4 FORCES AND EQUILIBRITJM^96
Problem 4.15.
(a) In Problem 4.14, if the magnitude of the applied force is F = 2.0 N, what is the magnitude and
direction of the frictional force on the book?
Solution
(a) Since the book is in equilibrium, the sum of the forces acting on it must equal zero. Figure 4-1 l(b) shows the body diagram for the book with all the forces acting on it. The fiictional force is f,, and the normal force is N. Noting that fs and F have no y components, from the condition that the sum
of the y components equals zero we have N - 10 N = 0, or N = 10 N.
(b) The maximum value attainable by the static friction force is /s,max = μsN = (0.25) (ION)= 2.5N
Fig. 4-
(a) (b)
F
F
Problem 4.14. A book of weight w = 10 N rests on a horizontal table top, as shown in Fig. 4-1 l(a),
and a horizontal force Fis applied to it. If the coefficient of static friction μs between the book and the
tabletop is 0.25, calculate (a) the normal force exerted by the tabletop on the book, and (b) the
maximum value of the static friction force.
static friction force ls.max that one surface can exert on another is proportional to the normal force N
between the surfaces: ls.max = μ)v, where N is the magnitude of the normal force, and μs is a
proportionally constant, called the coefficient of static friction, that depends on the nature of the two
surfaces. It is possible to force one object to slide over the other by applying a parallel force to one of
the objects that is larger than μsN, the maximum possible static friction force.
Fig. 4-
(Body diagrams)
CHAP 4] FORCES AND EQUILIBRlUM 97
T-fs=O or fs = T = ION
Since T <ls.max, the frictional force can balance T and the system remains at rest. The actual frictional
force can be obtained from the equilibrium of block A:
fs,max = μ 5 N = (0.5) (30N) = 15N
Problem 4.17. In Fig. 4-13(a), the two blocks are connected by a light rope over a frictionless,
weightless pulley. If the system is initially at rest, will it stay at rest? If so, what is the frictional force
exerted by the table on block A?
Solution Figure 4-13(b) gives the body diagrams for the two blocks. For block B, assuming equilibrium, the
y-component equation gives T - Wh = 0 or T = Wh = 10 N. Since we have a rope and a frictionless,
weightless pulley, the tension is the same on the block-A side of the pulley, and T = 10 N for block A as well. Vertical equilibrium of block A requires that N - Wa = 0, or N = Wa = 30 N. Then the maximum possible static frictional force is
1. The surface of the pulley is frictionless so that the cord slides effortlessly over it (frictionless
pulley).
2. The surface of the pulley has friction, but the pulley has no weight and there is no friction
between the pulley and the axle on which it rotates (weightless pulley).
In a problem, being told that the pulley is frictionless and/or weightless (massless) is generally
shorthand for case 1 or case 2, and you can assume as much unless told otherwise.
(b)^ (c) Fig. 4-
(a)
CHAP. 4] FORCES AND EQUILIBRIUM 99
Fig. 4-
x ------
y
I
I
I
° 18.4 (} = or i^ =^ !:^ I =^11 =^ tan (}
axis. x below the negative ° has magnitude 50.6 N and points away from the origin at 18.4 R Thus
makes with the R is the acute angle that If(} is in the third quadrant. R From the signs of its components,
axis, x negative
y=-(60N)sin37° 2 F F2x=-(60N)cos37° y=20N 1 F
(60N) (0.8) = -48N = 0 - Rx= Fix+ F2x
(60N) (0.6) = -16N = (20N) - F2y + F1y Ry=
= 50.6N 1/2 = [(-48)2+(-16)2] R
F1x=O
axes as shown in the figure and use the component method of y and x We choose 2.F + 1 F = R
addition.
of the two forces shown in Fig. 4-14. R Find the resultant Problem 4.18.
Solution
Problems for Review and Mind Stretching
Fig. 4- (b) (a)
N
t,
[CHAP. 4 FORCES AND EQUILIBRIUM^100
T
4-17 Fig. (b) (a)
Ti'
400N = 1 w
400 N hangs from a uniform heavy rope of length 3 m and = 1 w A block of weight Problem 4.22.
the force with which the rope pulls on the (a) 300 N, as shown in Fig. 4-17(a). Find = 2 w weight
the tension in the rope 1 m above the contact point with the block; (c) the force with which (b) block;
the ceiling pulls on the rope.
Solution
Fig. 4-l 7(b) we have the body diagrams for the block, the lower third of the rope, and the full In
rope, respectively. Each is in equilibrium, and the vector sum of the forces on each equals zero. Since the
direction, only the equilibrium condition in that direction need be applied. y forces are all in the
which can be replaced by the single resultant^3 F and^ 2, F^ 1, Here we have the concurrent forces F
Clearly, to as obtained in Problem 4.19. B, with line of action through point 3 F + 2 F + 1 F = R force
27. 7 N = E Thus R. = - E have equilibrium, the added fourth force, the equilibrant E, must obey
B. axis, with a line of action that must also pass through point x below the positive ° pointing 57.
Find the equilibrant of the forces in Problem 4.19. Problem 4.21.
Solution
4-16 Fig.
R=50.6N
E=50.6N
y
[CHAP. 4 FORCES AND EQUILIBRillM^102
Problem 4.24. A block of weight w = 200 N is pulled along a horizontal surface at constant speed
by a force F = 80 N acting at an angle of 30° above the horizontal, as shown in Fig. 4-19.
Finally, T 2 = I .532T3 = 467 N.
or l.970T3 = 600 N or T3 = 305N
,
T1y + T2y + T3y = -T, + T2 sin 60 ° + T3 cos 50 ° = 0 or 0.866T2 + 0.643T3 = 600N
Substituting for T2,
(0.866) (1.532T3) + 0.643T3 = 600N
Solution From the equilibrium of the block, T 1 = 600 N. Since the knot is in equilibrium, the body diagram, Fig. 4-18(b), gives T 1 + T 2 + T 3 = 0. Using the component method, we get Tix+ T2x + T3x = 0 - T2 cos60° + T3 sin50° = 0 or 0.5T2 = 0.766Tv or T 2 = 1.532 T3. (A sine appears in the x-component equation because the angle of T, is given relative to they axis). Similarly,
Fig. 4-
(a) (b)
L.
Problem 4.23. For the weight-and-strings setup of Fig. 4-18(a), find the tensions T1, T2, and T3.
(a) For the block, T 1 - w 1 = 0, or T 1 = 400 N equals the force of the rope on the block.
(b) For the lower third of the rope, T2 - Ti - w' = 0, where T 2 is the contact force of the upper two- thirds of the rope on the lower third and is the tension in the rope at that point; T{ is the force of the block on the rope, given by Newton's third law as T{ = T 1 = 400 N; w' is the weight of the lower third of the rope, or w' = 100 N. Thus T 2 = T{ + w' = 400 N + 100 N = 500 N.
(c) For the rope as a whole, T3 - T{ - w 2 = 0, or T3 = T{ + w 2 = 400 N + 300 N = 700 N, equals
the force of the ceiling on the rope.
CHAP. 4] FORCES AND EQUILIBRIUM 103
Problem 4.27. For the setup in Fig. 4-18(a)-first discussed in Problem 4.23-the breaking point of
the two cords attached to the wall and ceiling is 1500 N. How heavy can the block be without one of
the cords snapping? Assume the cord attached to the block can handle any weight.
T - μ~ - w 2 sin 3 7 ° = 0 and T = 300N + 120N = 420N
Finally, w 1 = T = 420 N.
Since the hanging block obeys w 1 = T, we have our result, w 1 -e 180 N.
(b) If the block is moving up the incline at constant speed, we proceed as before, noting that the frictional force is now directed down the incline although it still has the same magnitude fi = μkN. Furthermore the y-component equation for the block on the incline is unchanged, so we still have ·N = 400 N andfi = 0.3(400 N) = 120 N. The x-component equation changes only in that the sign of the x-component of the frictional force changes, and we get
T = (500N) (0.6) - 0.3(400N) = 300N - 120N = 180N
( Substituting into the previous equation we have
or N = (500N) (0.8) = 400N
For the y components
N -w2cos37° = 0
or T = (500N) (0.6) - 0.3N
Problem 4.26. Suppose that in Problem 4.25 there was friction between the block and the incline
and that the coefficient of sliding friction was μk = 0.3, but all the other data in the problem remained
unchanged. Find the weight of the hanging block, w1, if the other block moves at constant speed
(a) down the incline; (b) up the incline.
Solution
(a) We can use Fig~ 4-20 with the modification that there is an additional force on the block on the incline, a frictional force of magnitude fi opposing the motion of the block and hence pointing
parallel to the incline in the upward direction. From the rules for friction we have fk = μkN, where
N is the normal force exerted on the block by the incline. Following the reasoning of Problem 4. we now have for the x components
Then from our earlier result w 1 = T = 300 N. Note that we did not need to solve the y-component
equilibrium equation for the block on the incline to solve for Tand w1. This is because they-component
- equation gives us the normal force N, which does not affect the x-component equation when there is no friction. If the block were moving up the incline, the blocks would still be in equilibrium under the action of the same forces, so the answer would remain the same.
T-w2sin0=0 or T = (500N) (sin37°) = 300N
Solution In Fig. 4-20, all the forces on the respective blocks are shown right on the diagram for the system as a whole. Since both blocks move in straight lines at constant speed, they are each in equilibrium. For the
hanging block, using y components, we have T 1 - w 1 = 0, or w 1 = T1• To find T 1 we turn to the block
on the incline. We choose x and y axes along the incline and perpendicular to it, respectively. We also
note that the force of the cord on each block has the same magnitude, so T 2 = T 1 = T, since the cord is
light and the pulley is frictionless. Then, for the x-component equilibrium equation we get
CHAP. 4) FORCES AND EQUILIBRIUM 105
Fig. 4-22(a), assume the somewhat artificial condition that the strut is weightless and the In Problem 4.31. V
37 ° with the strut. = wall is frictionless. The cord makes an angle ()
if the strut is to be in translational equilibrium? N and T What are the conditions imposed on (a)
Can the strut be in rotational equilibrium under the circumstances shown? Give your justification. (b)
= 66 N; (b) No. The three forces cannot possibly be concurrent. N = 83 N, T (a) Ans.
2 F as shown in Fig. 4-21. What force 1 F A uniform rod of weight 100 N is acted on by a force Problem 4.30. V
must be added to the rod to ensure translational equilibrium?
axis x 56.6 N at an angle of 58.0 ° above the negative Ans.
automobile travels in a straight line with no skidding. An Problem 4.29.
the automobile travels at constant speed, are its wheels in translational and/or rotational equilibrium? If (a)
rotational translational and/or in are its wheels 60 mph, O to accelerates from automobile the If (b)
equilibrium?
Is a bit of chewing gum on the rim of a wheel of the automobile in translational and/or rotational (c)
equilibrium for the case of part (a) or part (b)?
translational equilibrium for either case; the bit of gum in (a) In both; (b) in neither; (c) not Ans.
goes through one rotation every time the wheel makes one complete turn. For part (a) it is in
it is not. b) rotational equilibrium, while for part (
The earth's moon revolves about the earth once a month and always keeps the same side facing the earth. (a)
Describe the translational and rotational motion of the moon.
Is the moon in translational and/or rotational equilibrium? (b)
The moon as a whole translates in a circular orbit about the earth; it rotates on its axis (a) Ans.
once a month.
the moon's monthly rotation on its axis if The moon is not in translational equilibrium; (b)
is uniform (it is, approximately), then the moon is in rotational equilibrium.
Problem 4.28.
Problems Supplementary
;'
1500 N; from = T2 will reach 1500 N first. We now set T2 and hence 2, is always less than T T3 Clearly
980 N. We can now determine the corresponding = 0.653 (1500 N) = 3 T above, this immediately yields
direction: y of the block using the equilibrium equation in the w weight
1929N = (980N) (0.643) + (1500N) (0.866) = cos50° T3 + sin60° T2 = 1 T = w
or 0.50T2 = 0.766T3or cos 60 ° T2 = sin 50 ° T
Solution
We first determine which of the two cords will reach a tension of 1500 N first. To do this we recall
direction requires x from Problem 4.23 that equilibrium in the
[CHAP. 4 FORCES AND EQUILIBRIUM 106
