Worked Examples-Calculation Part 2 - Equilibrium

Considered Formula;

$$\text {i)}\quad C.W.M = A.C.W.M$$ $$\qquad \text {Where}~~C.W.M \to Sum~of~clockwise~moment,\\\qquad \qquad \quad A.C.W.M \to Sum~of~Anti-clockwise~moment$$ $$\text {ii)}\quad F_u = F_d$$ $$\qquad \text {Where}~~F_u \to Sum~of~upward~forces,\\\qquad \qquad \quad F_d \to Sum~of~downward~forces$$

Example 1

A heavy uniform beam AB of weight 500N is supported at its ends. The beam carries a weight of 300N at a distance of 1.5m from the end A. if the beam is 4m long. Find the thrust/tension/reaction at A and B

Solution

Turning point at Point A

$$Sum~of~clockwise~moment = 3000(1.5)+ 500(2)$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 4500+1000$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 5500~Nm$$

$$Sum~of~Anti-clockwise~moment = Y(4)$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 4Y$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C.W.M = A.C.W.M$$ $$\qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~5500 = 4Y$$ $$\qquad ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac {5500}{4} = \frac {4Y}{4}$$ $$\qquad \quad~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Y = 1375~N$$

$$~~~~~~~~~~~~~~~~Sum~of~upward~forces = Sum~of~downward~forces$$ $$\qquad \qquad \qquad ~~~~~~~~~~~~~~~~~~~~X+Y = 3000+500$$ $$\qquad \qquad \qquad ~~~~~~~~~~~~~~~~X+1375 = 3500$$ $$\qquad \qquad \qquad \qquad \qquad ~~~~~~~~~~~~~X = 3500-1375$$ $$\qquad \qquad \qquad \qquad \qquad ~~~~~~~~~~~~~~~~~ = 2125~N$$

The reactions at A and B are 1375 N and 2125 N respectively