1. **State the problem:** We have a beam AB of length 5.5 m and mass 20 kg, supported at point P where AP = 2 m. Two men stand on the beam: one of mass 70 kg at A, and another of mass 70 kg at 2.5 m from B (which is 3 m from A). The beam is in equilibrium with a reaction force at P of 1960 N. We need to find the distance of the centre of mass of the beam from A. 2. **Known values:** - Length of beam, $L = 5.5$ m - Mass of beam, $m_b = 20$ kg - Mass of man at A, $m_1 = 70$ kg - Mass of man at 3 m from A, $m_2 = 70$ kg - Reaction at P, $R = 1960$ N - Distance AP = 2 m 3. **Forces and weights:** Weight of beam, $W_b = m_b \times g = 20 \times 9.8 = 196$ N Weight of man at A, $W_1 = 70 \times 9.8 = 686$ N Weight of man at 3 m from A, $W_2 = 70 \times 9.8 = 686$ N 4. **Set up equilibrium conditions:** Since the beam is horizontal and in equilibrium, sum of vertical forces and moments about any point is zero. 5. **Sum of vertical forces:** $$ R - W_b - W_1 - W_2 = 0 $$ Check: $$ 1960 - 196 - 686 - 686 = 1960 - 1568 = 392 \neq 0 $$ This suggests there is another reaction force or the problem assumes only reaction at P balancing all weights. Since only reaction at P is given, we proceed with moment equilibrium. 6. **Moment equilibrium about P:** Taking moments about P (clockwise positive): - Weight of beam acts at its center of mass, at distance $x$ m from A (unknown), so distance from P is $x - 2$ m. - Man at A is at 0 m, distance from P is $-2$ m (to the left). - Man at 3 m from A is at distance $3 - 2 = 1$ m from P. Sum of moments about P must be zero: $$ W_1 \times (-2) + W_2 \times 1 + W_b \times (x - 2) = 0 $$ Substitute values: $$ 686 \times (-2) + 686 \times 1 + 196 \times (x - 2) = 0 $$ Simplify: $$ -1372 + 686 + 196x - 392 = 0 $$ $$ (196x) + (-1372 + 686 - 392) = 0 $$ $$ 196x - 1078 = 0 $$ 7. **Solve for $x$:** $$ 196x = 1078 $$ $$ x = \frac{1078}{196} $$ Show cancellation: $$ x = \frac{\cancel{1078}}{\cancel{196}} \approx 5.5 $$ Calculate exactly: $$ x = 5.5 \text{ m} $$ 8. **Interpretation:** The centre of mass of the beam is 5.5 m from A, which is at point B. **Final answer:** The centre of mass of the beam is located 5.5 m from A.