Coplanar
forces are forces that lie in the some pane.
Parallel forces are forces whose tings of action are all parallel to
each other.

A body
acted upon by several forces said to be in equilibrium if it does not move or
rotate. Under this equilibrium
condition, the sum of the forces acting in one direction (e.g. upwards) must be
equal to the sum of the forces acting in opposite direction acting downwards
must balance the total prices acting downwards.

Also, the
body can only remain in equilibrium if the moments of the forces about any
point act, in such a way as to cancel
each other. That is, the total clockwise
moments of all the forces about any point of the object must be exactly counter
balanced by the total anti-clockwise moment about the same point.

Hence the
two conditions for equilibrium of parallel coplanar forces can be stated as
follows;

1) Forces:
The algebraic sum of the forces acting on the body in any given
direction must be zero. That is, the sum
of the upward forces must be equal the sum of the forces acting in one
direction must be equal to the sum of the forces acting opposite in direction.

2) Moments: The algebraic sum of the
moments of all forces about any point on the body must be zero or the total
clockwise moments of the forces about the same point. The second condition above is known as the
principle of moments.

The
principle of moments states that if a body is in equilibrium, then the sum of
the clockwise moments about any point on the body is equal to the sum of
anti-clock-wise moments about the same point.

**Example 1**

A light
beam AB sits on two pivots C and D.A load of 10N hangs at 0.2m from the support
act C. Find the value of the reaction
forces P and D at C and D. Take the
distance as shown below:

**Solution**

From 1

^{st}condition of equilibrium, total upward forces total downward forces (P+Q) = 10N = P+Q = 10N.
From second condition of equilibrium, total
clockwise moments = total anti-clockwise moments. Therefore taking moments about C we have:

10 x 2 =Q x)2+6) =3Q

Q
= 20 =2 .5N

8

Hence P =
10 –Q = 10 – 2.5 =7.5N

Alternatively,
we can take moment D

Hence;

P x 8 =
10 x6

P = 10 x 6 = 7.5N

8

Q = 10
-7.5 = 2.5N

Thus we
can take moment about C or D and get the same results.

**Example 2**

A pole AB
of length 10.0m and weight 1000N has its centre of gravity 4.0m from the end A,
and lies on horizontal ground. Draw a diagram to show the forces acting on the
pole when the end B is lifted by a vertical force. Calculate the force required to begin to lift
this end. Prove that this force applied
at the end A will not be sufficient to lift the end A from the ground.

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