• Dynamic equilibrium: object moves with constant velocity in a straight
line.
We note that F net and as are both vector quantities, and so in terms of their
components,
(Fnet )x =
X
(Fi )x = 0 ,
i
ax =
X
(ai )x = 0 ,
(4.11)
i
with similar equations for the y- and z-components.
Example: (Static equilibrium)
26
the three ropes in figure 36 are tied to
a small, very light ring. Two of the ropes are anchored to the walls at right
angles, and the third rope pulls as shown. What are T1 and T2 , the magnitudes
of the tension forces in the first two ropes?
Figure 36: Static equilibrium.
Solution: for the ring to be in static equilibrium, we must have, from Newton’s First Law, equation (4.11),
26
Knight, Exercise 1, page 146
56
(F net )x = (T 1 )x + (T 2 )x + (T 3 )x = 0
(4.12)
(F net )x = (T 1 )y + (T 2 )y + (T 3 )y = 0 .
(4.13)
and
The components of the force are:
x :
(T 1 )x = −T1 ,
y :
(T 1 )y = 0 ,
(T 2 )y = 0 ,
(T 2 )y = T2 ,
(T 3 )x = T3 cos 30◦
(T 3 )y = −T3 sin 30◦
Hence, using equations (4.12) and (4.13), we have
−T1 + 0 + T3 cos 30◦ = 0
−→
√
T1 = 100 cos 30◦ = 50 3 N ,
(4.14)
T2 = 100 sin 30◦ = 50 N ,
(4.15)
and
0 + T2 − T3 sin 30◦ = 0
−→
respectively, where we have used T3 = 100 N.
For an example concerning dynamic equilibrium, see Additional Materials.
4.8
27
Non-equilibrium Conditions
• The equilibrium conditions (4.9) and (4.10) are a special case of the more
general non-equilibrium situation where
27
Equilibrium Particle Dynamics Example, http://www.nanotech.uwaterloo.ca/∼ne131/
57
– The net force on an object is non-zero.
– The object accelerates and is no longer in equilibrium.
It is Newton’s 2nd Law which therefore provides the more general link between
force and motion: we write
F net =
X
F i = mas ,
X
(F i )x = max
(4.17)
X
(F i )y = may .
(4.18)
(4.16)
i
with the corresponding component equations being
(F net )x =
i
and
(F net )y =
i
Example:
28
in each of the diagrams in figure 37,
29
the forces are acting
on a 2.0 kg object. For each figure, find the values ax and ay , the x- and
y-components of the acceleration.
Solution: considering each diagram in turn:
• Diagram 1: the x- and y-components are as follows;
x : (F 1 )x = 4 N ,
(F 2 )x = 0 N ,
y : (F 1 )y = 3 N ,
(F 2 )y = −3 N ,
(F 3 )x = −2 N
(F 3 )y = 0 N .
Thus, substituting in equations (4.17) and (4.18) gives us
4 N − 0 N − 2 N = 2 ax
28
29
Knight, Exercise 5, page 146
Knight, Figures 5.5(a),(b), page 146
58
−→
ax = 1 m s−2 ,
Diagram 1
Diagram 2
Figure 37: Non-equilibrium motion.
and
3 N − 3 N + 0 N = 2 ay
−→
ay = 0 m s−2 ,
respectively.
• Diagram 2: the x- and y-components are:
x : (F 1 )x = 0 N, (F 2 )x = 4 N, (F 3 )x = 0 N, (F 4 )x = 0 N, (F 5 )x = −2 N
y : (F 1 )y = 3 N, (F 2 )y = 0 N, (F 3 )y = −2 N, (F 4 )y = −1 N, (F 5 )y = 0 N .
Substitution in equations (4.17) and (4.18) gives
0 N + 4 N + 0 N + 0 N − 2 N = 2 ax
−→
ax = 1 m s−2
3 N + 0 N − 2 N − 1 N + 0 N = 2 ay
−→
ay = 0 m s−2 ,
and
respectively.
59
4.9
Common Types of Forces – Revisited
Let us apply Newton’s laws to the forces discussed previously.
4.9.1
Weight
Recall that mass and weight are not the same thing:
• Mass is an intrinsic property of an object, indicating the amount of
matter it contains.
• Weight is the force on an object due to its presence in the gravitational
field of the Earth.
From equation (4.2), they are related via the expression
W = m afree fall ,
(4.19)
where W has the magnitude W = mg, and acts vertically downwards.
Equation (4.19) is a form of Newton’s 2nd Law; omitting the drag force due
to air resistance, an object in free fall experiences a net force
F net = W = m afree fall .
(4.20)
Apparent Weight We sense gravity indirectly through the normal force
due to the contact surface
• The weight force presses down on the ground, chair, etc.
• The ground, chair, etc. exerts a normal force upwards upon us.
Standing at rest on the ground (figure 38), we experience a net force of zero
vertically downwards:
Equilibrium: applying Newton’s 1st Law, equation (4.11),
(F net )y = N − W = 0 ,
60
i.e.
N = W = m afree fall = m g .
(4.21)
i.e.
in our inertial reference
frame, we sense our weight as
the normal force pushing up on
our feet with a magnitude mg.
Figure 38: Object at rest.
What happens when there is a resulting vertical force acting upon us? For
example, consider motion in an elevator
Upwards acceleration: for an acceleration as 6= 0 upwards (figure 39):
Non-equilibrium: applying Newton’s 2nd Law, equation (4.18):
(F net )y = N − W = m as ,
i.e.
N = W + mas .
(4.22)
Figure 39: Upwards acceleration in an elevator.
In our non-inertial reference frame accelerating upwards, the normal force pushing upwards on our feet - our sensation of weight has increased!
Note: it is important to realise that
61
• Our apparent weight W app – has increased.
• Our true weight W – is still the same.
Substituting for our true weight in equation (4.22) defines the magnitude of
our apparent weight to be the magnitude of the contact force N , i.e.
Wapp
as
as
=W 1+
= mg 1 +
g
g
(upwards acceleration) . (4.23)
So this explains why we feel heavier when an elevator suddenly accelerates
upwards.
Downwards acceleration: for an acceleration as 6= 0 downwards, where
as < g (figure 40):
Non-equilibrium: applying
Newton’s 2nd Law (4.18):
(F net )y = W − N = m as ,
i.e.
N = W − mas .
(4.24)
Figure 40: Downwards acceleration in an elevator.
In our non-inertial reference frame accelerating downwards, the normal force pushing upwards on our feet - our sensation of weight has decreased!
So by accelerating downwards in an elevator, we feel lighter than usual. By
analogy with equation (4.23), our apparent weight is
62
Wapp
as
= mg 1 −
g
=W
as
1−
g
(downwards acceleration, as < g) .
(4.25)
Apparent weight has several important applications: e.g. the upwards acceleration during a rocket launch is as g – under such conditions, equation
(4.23) shows us that astronauts will experience an extremely large apparent
weight.
Weightlessness In the elevator example, what happens if the cable snaps?
We will then be in free fall, i.e. mas = mg, and from equation (4.25), our
apparent weight will be
Wapp = 0 .
(4.26)
We lose our sensation of weight – there would be no normal force pushing on
our feet – and thus be weightless.
Note: although our apparent weight would be zero, our true weight is still of
magnitude mg.
4.9.2
Friction
Recall our classification of static, kinetic and rolling friction.
• Static friction f s : the frictional force on an object which prevents it
from slipping. Suppose we push an object with a horizontal force F push
(figure 41
30
)
The box remains at rest, so it is in static equilibrium; from Newton’s 1st
Law, we have
30
Knight, Figures 5.9 (a),(b), page 132
63
Example of static friction.
Force diagram for static friction.
Figure 41: Static friction.
fs = Fpush .
(4.27)
If Fpush is large enough, the object moves - this suggests the upper limit
fs max to fs :
– fs < fs max – object remains at rest.
– fs = fs max – object slips.
– fs > fs max – unphysical.
Furthermore, fs max is proportional to the magnitude of the normal force,
i.e. fs max ∝ N , or
fs max = µs N ,
(4.28)
where we have introduced the coefficient of static friction, µ s , a number which depends upon the materials making up the object and surface.
Note: in equation (4.28), we are saying that fs max = µs N , and not fs = µs N .
• Kinetic friction f k : the frictional force that appears when an object
moves across the surface. When the object begins to slide (figure 42
31
Knight, Figures 5.11 (a),(b), page 133
64
31
),