Monday, 30 May 2016

Conditions of equilibrium under the action of parallel coplanar forces



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 1st 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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