Chapter 4
Forces I
4.1 The Important Stuff
4.1 .1 Newton’s First Law
With Newton’s Laws we begin the study of how motion occurs in the real world. The study
of the causes of motion is called dynamics, or mechanics. The r elation between force
and acceleration was given by Isaac Newton in his three laws of motion, which form the
basis of elementary physics. Though Newton’s formulation of physics had to be replaced
later on to deal with motion at spe eds comparable to the spe ed light and for m otion on the
scale of atoms, it is applicable to everyday situations and is sti ll the be st introduction to
the f undamental laws of nature. The study of Newton’s laws and their implications is often
called Newtonian or classical mechanics.
Particles accelerate because they are being acted on by forces. In the absence of forces,
a particle will not accelerate, that is, it will move at constant veloc ity.
The user–friendly way of stating Newton’s First Law is:
Consider a body on which no force is acting. Then if it is at rest it
will remain at rest, and if it is moving with constant velocity it will
continue to move at that velo city.
Forces serve t o change the veloc ity of an object, not to maintain its motion (contrary to
the ideas of philosophers in ancient times).
4.1 .2 Newton’s Second Law
Experiments show that objects have a property called mass which measures how their motion
is influenced by forces. Mass is measured in kilograms in the SI system.
Newton’s Se cond Law is a relation between the net force (F) acting on a mass m and
its acceleration a. It says:
P
F = ma
77
78 CHAPTER 4. FORCES I
In words, Newton’s Second Law tells us to add up the forces acting on a mass m; this
sum,
P
F (or, F
net
) is equal to the mass m times i t s acceleration a.
This i s a vector relation; if we are working in two dimensions, this equation implies both
of the following:
X
F
x
= ma
x
and
X
F
y
= ma
y
(4.1)
The units of force must be kg ·
m
s
2
, which is abbreviated 1 newton (N), to honor Isaac
Newton (1642–1727), famous physicist and smart person. Thus:
1 newton = 1 N = 1
kg·m
s
2
(4.2)
Two other units of force which we encounte r sometimes are:
1 dyne = 1
g·cm
s
2
= 10
−5
N and 1 pound = 1 lb = 4.45 N
4.1 .3 Examples of Forces
To begin our study of dynamics we consider problems involving simpl e objects in sim ple
situations. Our first proble ms will involve little more than small masses, hard, smooth
surfaces and ideal strings, or objects that can be treated as such.
For all masses near the earth’s surface, the earth exerts a downward gravitational force
which is known as t he weight of the mass and has a magnitude given by
W = mg
A taught string (a string “under t ension”) exerts forces on the object s which are attached
to eithe r end. The forces are directed inward along the length of the string.) In our first
problems we will make the approximation that the string has no mass, and when it passes
over any pulley, the pulley’s mass can also be ignored. In that c ase, the magnitude of the
string’s force on e ither end is the same and will usually be called T , the string’s tension.
A solid surface will exert forces on a mass whi ch is in contact with it . In general the force
from the surface will have a perpendicular ( normal) component which we call the norm al
force of the surface. The surface can also exert a force which is parallel; this is a friction
force and will be covered in the next chapter.
4.1 .4 Newton’s Third Law
Consider two objects A and B. The force which object A exerts on
obje ct B is equal and opposite to the force which object B exerts on
obje ct A: F
AB
= −F
BA
This law is p opularly stated as the “law of action and reaction”, but in fact it de al s with
the forces betwe en two objects.
4.2. WOR KED EXAMPLES 79
4.1 .5 Applying Newton’s Laws
In this chapter we will look at some applications of Newton’s law to simple systems involving
small blocks, surfaces and strings. (In the next chapter we’ll deal with more compli cated
examples.)
A useful practice for problems involving more than one force is to draw a diagram showing
the individual masses in the problem along with the vectors showing the directions and
magnitudes of the individual forces. It is so important to do this that these diagrams are
given a special name, free–body diagrams.
4.2 Worked Examples
4.2 .1 Newton’s Second Law
1. A 3.0 kg mass undergoes an acceleration given by a = (2.0i + 5.0j)
m
s
2
. Find the
resultant force F and its magnitude. [Ser4 5-7 ]
Newton’s Se cond Law tells us that the resultant (net) force on a mass m is
P
F = ma.
So here we find:
F
net
= ma
= (3.0 kg)(2.0i + 5.0j)
m
s
2
= (6.0i + 15.j) N
The magnitude of the resultant force is
F
net
=
q
(6.0 N)
2
+ (15. N)
2
= 16. N
2. While two forces act on it, a particle of mass m = 3.2 kg is to move continuously
with velocity (3
m
s
)i − (4
m
s
)j. One of the forces is F
1
= (2 N)i + (−6 N)j. What is the
other force? [HRW5 5-5]
Newton’s Second Law tells us that if a is the acceleration of the particle , then (as there
are only two forces acting on it) we have:
F
1
+ F
2
= ma
But here the acceleration of the particle is z ero!! (Its velocity does not change.) This tells
us that
F
1
+ F
2
= 0 =⇒ F
2
= −F
1
80 CHAPTER 4. FORCES I
and so the second force is
F
2
= −F
1
= (−2 N)i + (6 N)j
This was a simple problem just to see if you’re paying attention!
3. A 4.0 kg object has a velocity of 3.0i
m
s
at one instant. Eight seconds later,
its velocity is (8.0i + 10.0j)
m
s
. Assuming t he object was subject to a constant net
force, find (a) the components of the force and (b) its magnitude. [Ser4 5-21]
(a) We are told that the (net) force acting on the mass was constant. Then we know that
its acceleration was also constant, and we can use the constant–acceleration results from
the previous chapter. We are given the initial and final velocities so we can compute the
components of the acceleration:
a
x
=
∆v
x
∆t
=
[(8.0
m
s
) − (3.0
m
s
)]
(8.0 s)
= 0.63
m
s
2
and
a
y
=
∆v
y
∆t
=
[(10.0
m
s
) − (0.0
m
s
)]
(8.0 s)
= 1.3
m
s
2
We have the mass of the obje ct, so from Newton’s Second Law we get t he c omponents of
the f orce:
F
x
= ma
x
= (4.0 kg)(0.63
m
s
2
) = 2.5 N
F
y
= ma
y
= (4.0 kg)(1.3
m
s
2
) = 5.0 N
(b) The magnitude of the (net) force is
F =
q
F
2
x
+ F
2
y
=
q
(2.5 N)
2
+ (5.0 N)
2
= 5.6 N
and its direction θ i s given by
tan θ =
F
y
F
x
=
5.0
2.5
= 2.0 =⇒ θ = tan
−1
(2.0) = 63.4
◦
(The question didn’t ask for the direction but there it is anyway!)
4. Five forces pull on the 4.0 kg box in Fig. 4.1. Find the box’s acceleration (a)
in unit–vector notation and (b) as a magnitude and direction. [HRW5 5-9]
(a) Newton’s Se cond Law will give the box’s acceler ation, but we must first find the sum of
the f orces on the box. A dding the x components of the forces gi ves:
X
F
x
= − 11 N + 14 N cos 30
◦
+ 3.0 N
= 4.1 N
4.2. WOR KED EXAMPLES 81
17 N
3.0 N
11 N
5.0 N
14 N
30
o
x
y
4.0 kg box
Figure 4.1: Five forces pull on a box in Example 4
(two of the forces have only y components). Adding the y components of the forces gives:
X
F
y
= +5.0 N + 14 N sin 30
◦
− 17 N
= − 5. 0 N
So the net force on the box (i n unit-vector notation) is
X
F = (4.1 N)i + (−5.0 N)j .
Then we find the x and y components of t he box’s acceleration using a =
P
F/m:
a
x
=
P
F
x
m
=
(4.1 N)
(4.0 kg)
= 1.0
m
s
2
a
y
=
P
F
y
m
=
(−5.0 N)
(4.0 kg)
= −1.2
m
s
2
So in unit–vector f orm, the accel eration of the box i s
a = (1.0
m
s
2
)i + (−1.2
m
s
2
)j
(b) The acceleration found in part (a) has a magnitude of
a =
q
a
2
x
+ a
2
y
=
q
(1.0
m
s
2
)
2
+ (−1.2
m
s
2
)
2
= 1.6
m
s
2
and to find i t s direction θ we calculate
tan θ =
a
y
a
x
=
−1.2
1.0
= −1.2
which gives us:
θ = tan
−1
(−1.2) = −50
◦
Here, since a
y
is negative and a
x
is positive, this choice for θ lies in the proper quadrant .
82 CHAPTER 4. FORCES I
4.2 .2 Examples of Forces
5. What are the weight in newtons and the mass in kilograms of (a) a 5.0 lb bag
of sugar, (b) a 240 lb fullback, and (c) a 1.8 ton automobile? (1 ton = 2000 lb.) [HRW5
5-13]
(a) The bag of sugar has a weight of 5.0 lb. (“Pound” is a unit of force, or weight.) Then
its weight i n newtons is
5.0 lb = (5.0 lb) ·
4.45 N
1 lb
= 22 N
Then from W = mg we calculate the mass of the bag,
m =
W
g
=
22 N
9.80
m
s
2
= 2.3 kg
(b) Similarly, the weight of t he fullback in newtons is
240 lb = (240 lb) ·
4.45 N
1 lb
= 1070 N
and then his (her) mass is
m =
W
g
=
1070 N
9.80
m
s
2
= 109 kg
(c) The automobile’s weight is given in tons; express it in newtons:
1.8 ton = (1.8 ton)
2000 lb
1 ton
!
4.45 N
1 lb
= 1.6 × 10
4
N .
Then i t s mass is
m =
W
g
=
1.6 × 10
4
N
9.80
m
s
2
= 1.6 × 10
3
kg
6. If a man weighs 875 N on Earth, what would he wei gh on Jupiter, where the
free–fall accele ration is 25.9
m
s
2
? [Ser4 5-12]
The weight of a mass m on the earth is W = mg where g is the free–fall acceleration on
Earth. The mass of the man is:
m =
W
g
=
875 N
9.80
m
s
2
= 89.3 kg
His weight on Jupiter is found using g
Jupiter
instead of g:
W
Jupiter
= mg
Jupiter
= (89.3 kg)(25.9
m
s
2
) = 2.31 × 10
3
N
The m an’s weight on Jupiter is 2.31 × 10
3
N.
(The statement of the problem is a little deceptive; Jupiter has no solid surface! The
planet will indeed pull on this man with a force of 2.31 × 10
3
N, but there is no “ground” to
push back!)
4.2. WOR KED EXAMPLES 83
40
o
50
o
60
o
T
1
T
2
T
1
T
2
5.0 kg
10 kg
(a)
(b)
T
3
T
3
Figure 4.2: Masses suspended by strings, for Example 7.
m
1
g
T
3
T
3
T
2
T
1
(a)
(b)
40
o
50
o
5.0 kg
Figure 4.3: Force diagrams for part (a) in Example 7.
4.2 .3 Applying Newton’s Laws
7. Find the tension in each cord for the systems shown i n Fig. 4.2. (Neglect the
mass of the cords.) [Ser4 5-26]
(a) In this part, we solve the system shown in Fig. 4.2(a).
Think of the forces acting on the 5.0 kg mass (which we’ll call m
1
). Gravity pulls down-
ward with a force of magnitude m
g
. The vertical string pulls upward with a force of magni-
tude T
3
. (These forces are shown in Fig. 4.3(a).) Since the hanging mass has no acceleration,
it must be true that T
3
= m
1
g. This gives us the val ue of T
3
:
T
3
= m
1
g = (5.0 kg)(9.80
m
s
2
) = 49 N .
Next we look at the forces which act at the point where all three strings join; these are
shown in Fig. 4.3(b). The force which the strings exert all point outward from the joining
point and from simple geometry they have the directions shown
Now this point is not accele r ating either, so the forces on it must all sum to zero. The
horizontal components and the vertic al components of these forces separately sum to ze ro.
84 CHAPTER 4. FORCES I
The horiz ontal components give:
−T
1
cos 40
◦
+ T
2
cos 50
◦
= 0
This equation by itself does not let us solve for the tensions, but it does give us:
T
2
cos 50
◦
= T
1
cos 40
◦
=⇒ T
2
=
cos 40
◦
cos 50
◦
T
2
= 1.19T
1
The vertical forces sum to zero, and thi s gives us:
T
1
sin 40
◦
+ T
2
sin 50
◦
− T
3
= 0
We already know the value of T
3
. Substituting this and also the expression for T
2
which we
just found, we get:
T
1
sin 40
◦
+ (1.19T
1
) sin 50
◦
− 49 N = 0
and now we can solve for T
1
. A little rearranging gives:
(1.56)T
1
= 49 N
which gives
T
1
=
49 N
(1.56)
= 31.5 N .
Now with T
1
in hand we get T
2
:
T
2
= (1.19)T
1
= (1.19)(31.5 N) = 37.5 N .
Summarizing, t he tensions i n the three strings for t his part of the problem are
T
1
= 31.5 N T
2
= 37.5 N T
3
= 49 N .
(b) Now we study the system shown in Fig. 4.2(b).
Once again, the net force on the hanging mass (which we call m
2
) must be zero. Since
gravity pulls down with a a force m
2
g and the verti cal string pulls upward with a force T
3
,
we know that we just have T
3
= m
2
g, so:
T
3
= m
2
g = (10 kg)(9.80
m
s
2
) = 98 N .
Now consider the forces which act at the place where all the strings meet. We do as in
part (a); the horizontal forces sum to zero, and this gives:
−T
1
cos 60
◦
+ T
2
= 0 =⇒ T
2
= T
1
cos 60
◦
The vertical forces sum to zero, giving us:
T
1
sin 60
◦
− T
3
= 0
4.2. WOR KED EXAMPLES 85
But notice that since we know T
3
, this equation has only one unknown. We find:
T
1
=
T
3
sin 60
◦
=
98 N
sin 60
◦
= 113 N .
Using this is our expression for T
2
gives:
T
2
= T
1
cos 60
◦
= (113 N) cos 60
◦
= 56.6 N
Summarizing, t he tensions in the three strings for this part of t he problem are
T
1
= 113 N T
2
= 56.6 N T
3
= 98 N .
8. A 210 kg mot orcycle accelerates from 0 to 55
mi
hr
in 6.0 s. (a) What is the
magnitude of the motorcycle’s constant acceleration? (b) What is the magnitude
of the net force causing t he acceleration? [HRW5 5-25]
(a) First, let’s convert some units:
55
mi
hr
= (55
mi
hr
)
1609 m
1 mi
1 hr
3600 s
!
= 24.6
m
s
so that the acceleration of the motorcycle is
a =
v
x
− v
x0
t
=
24.6
m
s
− 0
6.0 s
= 4.1
m
s
2
(b) Now that we know the acceleration of the m otorcycle (and its mass) we know the net
horizontal force, because N ewton’s Law tells us:
X
F
x
= ma
x
= (210 kg)(4.1
m
s
2
) = 8.6 × 10
2
N
The m agnitude of the net force on the motorcycl e is 8.6 × 10
2
N.
9. A rocket and its payload have a total mass of 5.0 × 10
4
kg. How l arge is the
force produced by the engine (the thrust) when (a) the rocket is “hovering” over
the launchpad just after ignition, and (b) when the rocket is accelerating upward
at 20
m
s
2
? [HRW5 5-35]
(a) First thing: draw a diagram of the forces acting on the rocket! This is done in Fig. 4.4. If
the mass of the rocket is M then we know that gravity will be exerting a force Mg downward.
The engine s (actually, the gas rushing out of the rocket) exerts a force of magnitude F
thrust
upward on the rocket.
If the rocket is hovering, i.e. it is motionless but off the ground then it has no acceleration;
so, here, a
y
=0. Newton’s Second Law then says:
X
F
y
= F
thrust
− Mg = Ma
y
= 0
86 CHAPTER 4. FORCES I
Acme
Rockets
Mg
F
thrust
y
Figure 4.4: Forces acting on the rocket in Example 9
F
q
30
o
(a)
(b)
mg
F
60
o
N
Figure 4.5: (a) Block held in place on a smooth ramp by a horizontal force. (b) Forces acting on the block.
which gives
F
thrust
= Mg = (5.0 × 10
4
kg)(9.80
m
s
2
) = 4.9 × 10
5
N
The e ngines exert an upward force of 4.9 × 10
5
N on the rocket.
(b) As in part (a), gravity and thrust are the only forces acting on the rocket, but now it
has an acceleration of a
y
= 20
m
s
2
. So Newton’s Second Law now gi ves
X
F
y
= F
thrust
− Mg = Ma
y
so that the force of the engines is
F
thrust
= Mg + Ma
y
= M( g + a
y
) = (5.0 × 10
4
kg)(9.80
m
s
2
+ 20
m
s
2
) = 1.5 × 10
6
N
10. A block of mass m = 2.0 kg is held in equilibrium on an incline of angle θ = 60
◦
by the horizontal force F, as shown in Fig. 4.5(a). (a) Det ermine the value of F ,
the magnit ude of F. (b) Determine the normal force exerted by the incline on
the block (ignore friction). [Se r4 5- 33]
4.2. WOR KED EXAMPLES 87
m
1
m
2
30
o
Figure 4.6: Masses m
1
and m
2
are connected by a cord; m
1
slides on frictionless slope.
(a) The first thing to do is to draw a diagram of the forces acting on the block, which we do
in F ig. 4.5(b). Gravity pulls downward with a force mg. The applied force, of magnitude F ,
is horizontal. The surface ex erts a normal force N on the block; using a little geometry, we
see that if the angle of the incline is 60
◦
, then the normal force is directed at 30
◦
above the
horizontal, as shown in Fig. 4.5(b). There is no f r iction force from the surface, so we have
shown all the forces acting on the block.
Oftentimes for problems involving a block on a slope it is easiest to use the components
of the gravity force along the slope and perpendicular to it. For this problem, this would
not make things any easier since the r e is no motion along the slope.
Now, the block is in equilibrium, meaning that it has no acceleration and the forces sum
to zero. The fact that the vertical components of the forces sum to zero gives us:
N sin 30
◦
− mg = 0 =⇒ N =
mg
sin 30
◦
Substitute and get:
N =
(2.0 kg)(9.80
m
s
2
)
sin 30
◦
= 39.2 N .
The horiz ontal forces also sum to zero, giving:
F − N cos 30
◦
= 0 =⇒ F = N cos 30
◦
= (39.2 N) cos 30
◦
= 33.9 N .
The appli ed force F is 33.9 N.
(b) The magnitude of the normal force was found in part (a); there we found:
N = 39.2 N .
11. A block of m ass m
1
= 3.70 kg on a frictionless inclined plane of angl e θ = 30.0
◦
is
connected by a cord over a massless, frictionless pulley to a second block of mass
m
2
= 2.30 kg hanging vertically, as shown in Fig. 4.6. What are (a) the magnitude
of the acceleration of each block and (b) the direction of the acceleration of m
2
?
(c) What is the tension in the cord? [HRW5 5-58]
88 CHAPTER 4. FORCES I
30
o
m
1
g sin q
T
N
x
y
m
1
g
m
1
g cos q
Figure 4.7: The forces acting on m
1
(a) Before thinking about the f orces acting on these blocks, we can think about their motion.
m
1
is c onstrained to move along the slope and m
2
must move vertically. Because the two
masses are joined by a string, the distance by which m
1
moves up the slope is equal t o the
distance which m
2
moves downward, and the amount by which m
1
moves down the slope
is the amount by which m
2
moves upward. The same is true of their accelerations; if it
turns out that m
1
is accelerating up the slope, that will be the same as m
2
’s downward
acceleration.
Now we draw “free–body diagrams” and invoke Newton’s Second Law for each mass.
Consider all the forces acting on m
1
. These are shown in Fig. 4.7
˙
The force of gravity, with
magnitude m
1
pulls straight down. Here, looking ahead to the fact that motion can only
occur along the slope it has decomposed into its components perpendicular to the surface
(with magnitude m
1
cos θ) and down the slope (with magnitude m
1
sin θ). The normal force
of the surface has magnitude N and points... normal to the surface ! Finally the string pulls
up with slope with a force of magnitude T , the tensi on in the string.
Suppose we let x be a coordinate which measures movement up the slop e. (Note, we are
not saying that the block will move up the slope, this is just a choice o f coordinate. Let y be
a co ordinate going perpendicular to the sl ope. We know that there is no y acceleration so
the components of the forces in the y direction must add to zero. Thi s gives:
N − m
1
g cos θ = 0 =⇒ N = m
1
g cos θ
which gives the normal force should we ever need it. (We won’t.) Next, the sum of t he x
forces gives m
1
a
x
, which will not be zero. We get:
T − m
1
g sin θ = m
1
a
x
(4.3)
Here there are two unknowns, T and a
x
.
The free–bo dy diagram for mass m
2
is shown in Fig. 4.8. The force of gravity, m
2
g
pulls downward and the string tension T pulls upward. Suppose we use a coordinate x
0
which points straight down. (This is a lit tle unconventional, but you can see that there is a
4.2. WOR KED EXAMPLES 89
T
m
2
g
x
Figure 4.8: The forces a cting on m
2
. Coordinate x
0
points downward.
connection with the coordinate x used for the motion of m
1
. Then the sum of f orces in the
x
0
direct ion gives m
2
a
x
0
:
m
2
g − T = m
2
a
x
0
Now as we argued above, the accelerations are equal: a
x
= a
x
0
. This gives us:
m
2
g − T = m
2
a
x
(4.4)
At thi s point, the physics is done and the rest of the problem is doing the math (al gebra)
to solve for a
x
and T . We are first interested in finding a
x
. We note that by adding Eqs. 4.3
and 4.4 we will eliminate T . Doing this, we get:
(T − m
1
g sin θ) + (m
2
g − T ) = m
1
a
x
+ m
2
a
x
this gives:
m
2
g − m
1
g sin θ = (m
1
+ m
2
)a
x
and finally:
a
x
=
(m
2
− m
1
g sin θ)g
m
1
+ m
2
Substituting the given values, we have:
a
x
=
(2.30 kg − 3.70 kg sin 30
◦
)(9.80
m
s
2
)
(3.70 kg + 2.30 kg)
= +0.735
m
s
2
So a
x
= +0.735
m
s
2
. What does this mean? It means that the acceleration of m
1
up the slope
and m
2
downwards has magnitude 0.735
m
s
2
. The plus sign in our result for a
x
is telling us
that the acceleration does go in the way we (arbitrarily) se t up the coordinates. If we had
made the opposite (“wrong”) choice for the coordinates then our acceleration would have
come out with a minus sign.
(b) We’ve already found the answer to this part in our understanding of the result for part
(a). Mass m
1
accelerates up the slope; mass m
2
accelerates verticall y downward.
90 CHAPTER 4. FORCES I
Acme
Bananas
Figure 4.9: Monkey runs up the rope in Example 12.
(c) To get the tension in the string we may use either Eq. 4.3 or Eq. 4.4. Using 4.4 gives:
m
2
g − T = m
2
a
x
=⇒ T = m
2
g − m
2
a
x
= m
2
(g − a
x
)
Substituting everything,
T = (2.30 kg)(9.80
m
s
2
− (0.735
m
s
2
)) = 20.8 N
12. A 10 kg monkey climbs up a massless rope that runs over a frictionless tree
limb (!) and back down to a 15 kg package on the ground, as shown in Fig. 4.9.
(a) What is the magnitude of the least accele ration the monkey must have if
it i s to lift the package off the ground? If, after the package has b een lifted
the monkey stops its climb and holds onto the rope, what are (b) the monkey’s
acceleration and (c) the tensi on in the rope? [HRW5 5-64]
(a) Before we do anything else, let’s understand what forces are acting on the two masses
in this problem. The free–body diagrams are shown in F ig. 4.10. The monkey holds onto
the rope so it exerts an upward force of magnitude T , where T is the tension in the rope.
Gravity pulls down on the monkey with a force of magnitude mg, where m is the mass of
the monkey. These are all the forces. Note that they will not cancel since the problem talks
about the monkey having an acceleration and so the net force on the monkey will not be
zero.
The forces acting on the box are also shown. Gravity pulls downward on the b ox with a
force of magnitude Mg, M being the mass of the box. The rope pulls upward wi th a force
T , If the box is resting on the ground, the ground wil l be pushing upward with some force
F
ground
. (Her e, the ground cannot pull downward.) However when the box is not touching
the ground then F
ground
will be zero.
4.2. WOR KED EXAMPLES 91
T
mg
T
Fground
Monkey
Box
Mg
Figure 4.10: The forces acting on the two masses in Example 12.
In the first part of the problem, the monkey is moving along the rope. It is not stuck
to any point of the rope, so there is no obvious relation between the acceleration of the
monkey and the acceleration of the box. Suppose we let a
y,monkey
be the vertical acceleration
of the monkey and a
y,box
be the vertical acceleration of the box. Then from our free–body
diagrams, Newton’s Second Law gives the acceleration of the monkey:
T − mg = ma
y,monkey
When the box is on the ground its acceleration is zero and then T + F
gr
= mg. But when
the box is off then ground then we have:
T − Mg = Ma
box
(Box off the ground)
In t he first part of the problem we are solving for the condition that the m onkey climbs
just barely fast enough for the box to be lifted off the ground. If so, then the ground would
exert no force but the net force on the box would be so sm al l as to be virtually zero; the box
has a very, very tiny accelerati on upwards. From this we k now:
T − Mg = 0 =⇒ T = Mg
and substituting this result into the first equation gives
Mg − mg = ma
monkey
=⇒ a
monkey
=
(M − m)g
m
Substituting the given values,
a
monkey
=
(15 kg − 10 kg)(9.80
m
s
2
)
10 kg
= 4.9
m
s
2
The m onkey must pull himself upwards so as to give himself an acce leration of 4.9
m
s
2
. Any-
thing l ess and the box wil l remain on the ground.
(b) Next, suppose that after climbing for while (during which time the box has been rising
off the ground) the monkey grabs onto the rope. What new condition does this gi ve us?
92 CHAPTER 4. FORCES I
Now it is true that the distance that the monkey moves up is the same as the distance which
the box moves down. The same is true of the velocities and acce lerations of the monkey and
box, so i n this part of the problem (recalling that I defined both accelerations as being in
the upward sense),
a
monkey
= −a
box
.
This condition is not true in general, but here it is because we are told that the monkey is
holding fast to the rope.
If you recal l the example of the Atwood machine from your textbook or lecture notes,
this is the same physical situation we are de al ing with here. We expect the less massive
monkey to accelerate upwards and the more massive box to accelerate downwards. Let’s use
the symb ol a for the monkey’s vertical acceleration; then the box’s vertical acceleration is
−a and our equations are:
T − mg = ma
and
T − Mg = M(−a) .
At this point the physics is done and the rest is math (algebra) to solve for the two unknowns,
T and a. Since the first of the se eq uations gives T = mg + ma, substituting this into the
second equation gives:
mg + ma − Mg = −Ma =⇒ ma + Ma = Mg − mg
which gives:
(M + m )a = (M − m)g =⇒ a =
(M − m)
(M + m )
g
Plugging in the numbers gives
a =
(15.0 kg − 10.0 kg)
(15.0 kg + 10.0 kg)
(9.80
m
s
2
) = 2.0
m
s
2
.
When the monkey is holding tight to the rope and b oth masses move freely, the monkey’s
acceleration is 2.0
m
s
2
upwards.
(c) Now that we have the acceler ation a for this part of the problem, we can easil y substitute
into our results in part (b) and find the tension T . From T − mg = ma we get:
T = mg + ma = m( g + a) = ( 10.0 kg)(9.80
m
s
2
+ 2.0
m
s
2
) = 118 N .
The te nsion in the rope is 118 N.
13. A mass M is held in place by an applied force F and a pulley system as shown
in Fig. 4.11. The pulleys are massless and frictionless. Find (a) the tension in
each section of rope, T
1
, T
2
, T
3
, T
4
, and T
5
, and (b) the m agni tude of F. [Ser4 5-65]
4.2. WOR KED EXAMPLES 93
T
1
T
2
T
3
T
4
T
5
M
F
Figure 4.11: Crudely-drawn hand supports a mass M by means of a rope and pulleys
(a) We note first that the m ass M (and therefore everything else) is motionless. This simpli-
fies the problem considerably! In particular, every mass in this problem has no acceleration
and so the total force on each mass is zero.
We have five r ope tensions to find here, so we’d better start writing down some equations,
fast! Actually, a few of them don’t take much work at all; we know that when we have the
(idealized) situation of massless rope passing around a frictionless massless pulley, the string
tension is the same on both sides. As shown in the figure, it is a single piece of rope that
wraps around the bi g upp er pulley and the lower one, so the tensions T
1
, T
2
and T
3
must be
the s ame:
T
1
= T
2
= T
3
Not bad so far!
Next , think about the forces acting on mass M. This is pretty simple... the force of
gravity Mg pulls down, and the tension T
5
pulls upward. That’s all the forces but they sum
to zero because M is motionless. So we must have
T
5
= Mg .
Next , consider the forces which act on the small pulley. These are diagrammed in
Fig. 4.12(a). There is a downward pull of magnitude T
5
from the rope which is attached
to M and also upward pulls of magnitude T
2
and T
3
from the long rop e which is wrapped
around the pulley. These forces must sum to zero, so
T
2
+ T
3
− T
5
= 0
But we al r eady know that T
5
= Mg and t hat T
2
= T
3
so this tells us that
2T
2
− Mg = 0
which gives
T
2
=
Mg
2
=⇒ T
3
= T
2
=
Mg
2
.
94 CHAPTER 4. FORCES I
(a)
(b)
T
5
T
3
T
2
T
1
T
2
T
3
T
4
Figure 4.12: (a) Forces on the small (lower) pulley. (b) Forces on the large (upper) pulley.
We also have: T
1
= T
2
= Mg/2.
Next , consider the forces on the large pulley, shown in Fig. 4.12(b). Tension T
4
(in the
rope attached to the ceiling) pulls upward and tensions T
1
, T
2
and T
3
pull downward. These
forces sum to zero, so
T
4
− T
1
− T
2
− T
3
= 0 .
But T
4
is the only unknown in this equation. Using our previous answers,
T
4
= T
1
+ T
2
+ T
3
=
Mg
2
+
Mg
2
+
Mg
2
=
3Mg
2
and so the answers are:
T
1
= T
2
= T
3
=
Mg
2
T
4
=
3Mg
2
T
5
= Mg
(b) The force F is the (downward) force of the hand on the rope. It has the same magnitude
as the force of t he rope on the hand, which is T
1
, and we found thi s to be Mg/2. So
F = Mg/2.
14. Mass m
1
on a frictionless horizontal table is connected to mass m
2
through
a massless pulley P
1
and a massless fixed pulley P
2
as shown in Fig. 4.13. (a)
If a
1
and a
2
are the magnitudes of the accelerations of m
1
and m
2
respectively,
what is the relationship between these accelerations? Find expressions for (b)
the tensions in the strings and (c) the accelerations a
1
and a
2
in terms of m
1
, m
2
and g. [Ser4 5-46]
(a) Clearly, as m
2
falls, m
1
will move to the right, pulled by the top string. But how do the
magnitudes of t he displacements, velocities and accelerations of m
2
and m
1
compare? They
are not necessarily t he same. Indeed, they are not the same.
Possibly the best way to show the relation between a
1
and a
2
is to do a li t t le math; for
a very complicated system we would have to do this anyway, and the practice won’t hurt.
4.2. WOR KED EXAMPLES 95
m
1
m
2
P
1
P
2
Figure 4.13: System of masses and pulleys for Example 14.
x
l
x
block
To hanging mass
Figure 4.14: Some geometry for Example 14.
96 CHAPTER 4. FORCES I
(a)
(b)
(c)
T
1
N
T
1
T
1
m
2
g
T
2
T
2
m
1
g
Figure 4.15: Forces on the masses (and moving pulley) in Exa mple 14. (a) Forces on m
1
. (b) Forces on
the moving (massless) pulley. (c) Forces on m
2
.
Focus on the uppe r mass (m
1
) and pulley P
1
, and consider the lengths labelled in Fig. 4.14.
x measures the distance from t he wall to the moving pulley; cl early the position of m
2
is also
measured by x. ` is the length of string f r om m
1
to the pulley. x
block
measures the distance
from the wall to m
1
. Then:
x
block
= x − ` .
This really ignores the bit of string that wraps around the pulley, but we can see that it
won’t m atter .
Now the total length of t he stri ng is L = x + ` and it does not change with t ime. Since
we have ` = L − x, we can rewrite the last equation as
x
block
= x − (L − x) = 2x − L
Take a couple time derivatives of this, keeping in mind that L is a constant. We get:
d
2
x
block
dt
2
= 2
d
2
x
dt
2
But the left side of this e quation is the acceleration of m
1
and the right side is the (magnitude
of the) acceleration of m
2
. The accel eration of m
1
is twice that of m
2
:
a
1
= 2a
2
We can also understand this result by realizing that when m
2
moves down by a distance
x, a l ength 2x of the string must go from the “underneath” section to the “above” section in
Fig. 4.14. Mass m
1
follows the end of the string so it must move forward by a distance 2x.
Its displacement is always twice that of m
2
so its acceleration is always twice that of m
2
.
(b) Now we try to get some information on the forces and accelerations, and we need to draw
free–body diagrams. We do this in Fig. 4.15. Mass m
1
has forces m
1
g acting downward, a
normal force from the table N acting upward, and the string tension T
1
pulling to the right.
The vertical forces cancel since m
1
has only a horizontal acc eleration, a
1
. Newton’s Second
Law gives us:
T
1
= m
1
a
1
(4.5)
4.2. WOR KED EXAMPLES 97
The pulley has forces acting on it, as shown i n Fig. 4.15(b). The string wrapped around
it exerts its pull (of magnitude T
1
) both at the top and bottom so we have two forces of
magnitude T
1
pulling to the left. The second string, which has a tension T
2
, pulls to the
right with a force of magnitude T
2
.
Now this is a slightly subtle point, but the forces on the pulley must add to zero because
the pulley i s assumed to be massless. (A net force on it would give it an infinite acceleration!)
This condition gives us:
T
2
− 2T
1
= 0 (4.6)
Lastly, we come to m
2
. It will accelerate downward with acceleration a
2
. Summing the
downward forces, Newton’s Second Law gives us:
m
2
g − T
2
= m
2
a
2
(4.7)
For goo d measure, we repeat the result found in part (a):
a
1
= 2a
2
(4.8)
In these equations, the unknowns are T
1
, T
2
, a
1
and a
2
. . . four of them. And we have
four equations relating the m, namely Eqs. 4.5 through 4.8. The physics is done. We just do
algebra to finish up the problem.
There are many ways to do the algebra, but I’ll grind through it in following way: Sub-
stitute Eq. 4.8 into Eq. 4.5 and get:
T
1
= 2m
1
a
2
Putting this result into Eq. 4.6 gives
T
2
− 2T
1
= T
2
− 4m
1
a
2
= 0 =⇒ T
2
= 4m
1
a
2
and finally using this in Eq. 4.7 gives
m
2
g − 4m
1
a
2
= m
2
a
2
at which point we can solve for a
2
we find:
m
2
g = a
2
(4m
1
+ m
2
) =⇒ a
2
=
m
2
g
(4m
1
+ m
2
)
(4.9)
Having solved for one of the unknowns we can quickly find the rest. Eq. 4.8 gives us a
1
:
a
1
= 2a
2
=
2m
2
g
(4m
1
+ m
2
)
(4.10)
Then Eq. 4.8 gives us T
1
:
T
1
= m
1
a
1
=
2m
1
m
2
g
(4m
1
+ m
2
)
(4.11)
98 CHAPTER 4. FORCES I
Finally, since Eq. 4.6 tells us that T
2
= 2T
1
we get
T
2
=
4m
1
m
2
g
(4m
1
+ m
2
)
(4.12)
Summarizing our results from Eqs. 4.9 through 4.12, we have:
T
1
=
2m
1
m
2
g
(4m
1
+ m
2
)
T
2
=
4m
1
m
2
g
(4m
1
+ m
2
)
for the tensions in the two strings and:
(c)
a
1
=
2m
2
g
(4m
1
+ m
2
)
a
2
=
m
2
g
(4m
1
+ m
2
)
for the accelerations of the two masses.
