In general, a body is subjected to sufficiently good numbers of forces. Consider for example, a book placed on a table. This simple arrangement actually is subjected to four normal forces operating at the four corners of the table top in the vertically upward direction and two weights, that of the book and the table, acting vertically downward.

If we want to solve for the four unknown normal forces acting on the corners of the table top, we would need to have a minimum of four equations. Clearly, two vector relations available for equilibrium are insufficient to deal with the situation.

We actually need to write two vector equations in component form along each of thee mutually perpendicular directions of a rectangular coordinate system. This gives us a set of six equations, enabling us to solve for the unknowns. We shall, however, see that this improvisation, though, helps us a great deal in analyzing equilibrium, but is not good enough for this particular case of the book and table arrangement. We shall explain this aspect in a separate section at the end of this module. Nevertheless, the component force and torque equations are :

∑
F
x
=
0
;
Σ
F
y
=
0
;
Σ
F
z
=
0
∑
F
x
=
0
;
Σ
F
y
=
0
;
Σ
F
z
=
0

∑
τ
x
=
0
;
Σ
τ
y
=
0
;
Σ
τ
z
=
0
∑
τ
x
=
0
;
Σ
τ
y
=
0
;
Σ
τ
z
=
0

The first set of equations are called “force balance equations”, whereas second set of equations are called “torque balance equations”.

We must understand that mere existence of component equations does not guarantee that we would have all six equations to work with. The availability of relations depend on the orientation of force system operating on the body and dimensions of force – whether it is in one, two or three dimensions. To understand this aspect, let us consider the case of static equilibrium of the beam as shown in the figure. The forces acting on the beam are coplanar in xy- plane.

Coplanar forces can rotate rigid body only around an axis perpendicular to the plane of forces. This is z-direction in this case. We can not write equations for torque around “x” or “y” axes. Thus, we have only torque balance equation in z-direction :

Σ
τ
z
=
0
Σ
τ
z
=
0

Further, all available balance equations may not be fruitful. Since forces lie in xy-plane in the example here, we would expect two force balance equations :

Σ
F
x
=
0
Σ
F
x
=
0

Σ
F
y
=
N
1
+
N
2
−
m
g
−
m
1
g
−
m
2
g
Σ
F
y
=
N
1
+
N
2
−
m
g
−
m
1
g
−
m
2
g

Clearly, forces have no components in x-direction. As such the force balance equation in x-direction is not meaningful.

We should apply available meaningful balance equations with certain well thought out plan. The most important is to select the point/ axis about which we calculate torque. In general, we choose the coordinate system in such a manner that moment arms of the most numbers of unknown forces become zero. This technique has the advantage that the resulting equation has least numbers of unknown and is likely to yield solution. In the figure shown here, the torque about a point “O” lying on the line of action of force “
N
1
N
1
” results in zero moment of arm for this force.

If “a”, “b”, “c” and “d” be the linear distances of force "
m
1
g
m
1
g
", mg, "
m
2
g
m
2
g
" and "
N
2
N
2
" respectively from z-axis, then considering anticlockwise torque positive and anti-clockwise torque negative, we have :

Σ
τ
z
=
0
X
N
1
−
a
X
m
1
g
−
b
X
m
g
−
c
X
m
2
g
+
d
X
N
2
Σ
τ
z
=
0
X
N
1
−
a
X
m
1
g
−
b
X
m
g
−
c
X
m
2
g
+
d
X
N
2

We can see that this equation eliminates “
N
1
N
1
” and as such reduces numbers of unknowns in the equation. We must, however, understand that the selection of appropriate axis for calculation of torque is critical to evaluate the unknown. For example, if we consider to calculate torque through a perpendicular axis passing through center of mass of the beam, then we loose a known quantity (mg) and include additional unknown quanity, "
N
2
N
2
" in the equation.

Problem 1 :
A uniform beam of length 20 m and linear mass density 0.1 kg/m rests horizontally on two pivots at its end. A block of mass of 4 kg rests on the beam with its center at a distance 5 m from the left end. Find the forces at the two ends of the beam.

Solution : The beam and block system is in static equilibrium. In order to write force balance equation, we first need to draw the free body diagram of the beam.

The normal forces at the ends act in vertically upward direction. Weights of the beam and block, on the other hand, acts downward through the corresponding centers of mass. Clearly, all forces are acting in vertical direction. As such, we can have one force balance equation in vertical direction. But, there are two unknowns. We, therefore, need to write torque balance equation about an axis perpendicular to the plane of figure (z-axis). The important aspect of choosing an axis is here that we should aim to choose the axis such that moment arm of one of the unknown forces is zero. This eliminates one of the forces from the torque balance equation.

Force balance equation in vertical direction is :

Σ
F
y
=
N
1
+
N
2
−
m
1
g
−
m
2
g
=
0
Σ
F
y
=
N
1
+
N
2
−
m
1
g
−
m
2
g
=
0

Where "
m
1
m
1
" and "
m
2
m
2
" are the masses of beam and block respectively. According to question,

m
1
=
λ
L
=
0.1
X
20
=
2
k
g
m
1
=
λ
L
=
0.1
X
20
=
2
k
g

m
2
=
4
k
g
m
2
=
4
k
g

Putting values,

⇒
N
1
+
N
2
−
2
X
10
−
4
X
10
=
0
⇒
N
1
+
N
2
−
2
X
10
−
4
X
10
=
0

Considering anticlockwise torque positive and clockwise torque negative, the torque balance equation is :

Σ
τ
z
=
0
X
N
1
+
L
X
N
2
−
L
1
X
m
1
−
L
2
X
m
2
=
0
Σ
τ
z
=
0
X
N
1
+
L
X
N
2
−
L
1
X
m
1
−
L
2
X
m
2
=
0

where “L”, “
L
1
L
1
” and “
L
2
L
2
” are moment arms of normal force “
N
2
N
2
”, weight of beam, “
m
1
g
m
1
g
” and weight of block, “
m
2
g
m
2
g
” respectively. Putting values,

⇒
20
X
N
2
−
10
X
2
X
10
−
5
X
4
X
10
=
0
⇒
20
X
N
2
−
10
X
2
X
10
−
5
X
4
X
10
=
0

⇒
N
2
=
20
N
⇒
N
2
=
20
N

Since one of the unknown normal forces is, now, known, we can solve force balance equation by substituting this value,

⇒
N
1
+
20
−
2
X
10
−
4
X
10
=
0
⇒
N
1
+
20
−
2
X
10
−
4
X
10
=
0

⇒
N
1
=
40
N
⇒
N
1
=
40
N