12.1
STATIC EQUILIBRIUM
EXERCISES
Section 12.1 Conditions for Equilibrium
14.
(a)
∑= + − − +
=
v
Fi
j
i
j
j
i
(
ˆˆˆˆ
ˆ
).
2223
0
N
(b)
()
ˆ
(
ˆˆ
)(
ˆ
)
∑=
×
++
−
×
−
−+
−
r
τ
i
iij
i
ij
0
22
2
2
3
7
[(
ˆ
ˆ
)
ˆ
]
ij j
+×
⋅=
Nm
ˆ
.
437
0
+−
⋅ =
k
15.
INTERPRET
We have been told that the choice of pivot point does not matter if the sum of forces is zero. Here we
will show that this is true for two different pivot points. Three forces are acting on an object, which is in equilibrium.
DEVELOP
The three forces are
r
Fij
1
=+
N at point
(,) (
)
,
xy
=
2m
,0m
r
j
2
23
=− −
N at
,
−
1m
and
r
Fj
3
1
=
ˆ
N at
.
−
7m
,1m
We find the torques due to these three forces around points
3m
,2m and
,
−
using
r
r
r
τ
=×

.
rF
To find the value of
r
r
for a point other than the origin, we take the vector difference between the
point where the force is applied and the point used as the pivot:
rr
r
r
=−
.
applied
pivot
EVALUATE
For point
,
,2m
the torque due to
F
1
is
r
r
11
02
×
=
−
+
−
(
(
)
ˆ
rrF
i
applied
pivot
m
ˆ
)
ˆ
(
ji
j
k
mN
Nm
1
×+
= − −
120
220
2
r
)
−=
42
kk
Nm
Nm
Similarly,
r
2
8
=
ˆ
k
Nm and
r
3
10
ˆ
k
Nm. The sum of these three is
r
total
k
=+−
=
ˆ
.
2
8
10
0
Nm
Around point (–7 m, 1 m), the torques are
r
1
20
=
ˆ
k
Nm,
r
2
20
ˆ
k
Nm, and
r
3
0
=
. The sum of these three is
also
r
total
=
0.
ASSESS
Note that the torque due to force 3 around the second pivot is zero, since the force acts on the pivot.
16.
The conditions for static equilibrium, under the action of three forces, can be written as:
vv
v
FF
F
31
2
−
and
v
v
v
v
v
v
rFrF
33
×=
−×+×
.
(a)
In this case,
v
j
r
j
i
2
==
=
ˆ
,(
,
m) ,
2
and
v
rj
2
1
=
ˆ
.
m
Thus,
v
i
j
3
+
(
),
which is a force of magnitude
24
5
F
,
°
down into the third quadrant (
θ
x
=°
−
°
225
135
or
CCW from the
x
axis).
The point of application,
v
r
3
, can be found from the second condition,
v
v
rF x
iy
jF
iF
j
3
3
×= + − − =
(
)
ˆ
=
−×−× =−
× =
F
k rFrF
i
1122
01
v
v
v
v
m(
)
ˆ
.
Fk
Thus,
−+ =
1m, or the line of action of
v
F
3
passes through the point of application of
v
F
2
(the point ( ,
)).
01m
Any point on this line is a suitable point of
application for
v
F
3
(e.g., the point ( ,
(b)
In this case,
12
so
v
F
3
0
=
, but
v
v
v
v
112 2
×+×=
v
rr F
21
2
0
−×≠
so
v
v
0
×≠
. Thus there is no single force that can be added to produce static equilibrium.
17.
INTERPRET
The problem asks for a set of conditions that must be met for the body to be in static equilibrium.
This means that all the external forces and torques must be zero.
DEVELOP
Static equilibrium demands that (see Equations 12.1 and 12.2)
Σ
ΣΣ
r
r
r
r
F
i
ii
i
=
=
0
0
12
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View Full Document12.2
Chapter 12
Since all of the forces lie in the same plane, which includes the points
O
and
P
, there are two independent
components of the force condition (Equation 12.1) and one component of the torque condition (Equation 12.2).
EVALUATE
(a)
Taking the
x
axis to the right, the
y
axis up and the
z
axis out of the page in Figure 12.12, we have:
0
0
0
12
3
24
=∑
=− +
+
=−
+
FF
F
F
F
x
y
zO
sin
cos
()
φ
τ
−
+
LF
23
sin
cos
φφ
(b)
The equation for torque about point
P
is
0
21
2
1 2
+
−
s
i
n(
)
τφ
zP
L
L F
The lever arms of all the forces about either
O
or
P
should be evident from Fig. 12.12.
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 Fall '10
 Pheong
 Force, Static Equilibrium, µs

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