Non-uniform Beam on Two Supports
A non-uniform horizontal beam AB has length 6.0 m and mass 25 kg. Its centre of mass is 2.5 m from A. The beam is supported by a pivot at A and by a roller at C, where C is 4.5 m from A. A load of mass 20 kg hangs from B. The system is in equilibrium and the beam is horizontal. Take g = 9.81 m/s². The roller can exert only an upward force, while the pivot at A can exert either an upward or a downward vertical force. Friction at the pivot is negligible, and the beam remains straight. You are asked to find the reaction forces at A and C, to determine the greatest load that can be hung at B before the beam is on the point of tipping about C, and to comment on what happens if that load is exceeded.
1. Calculate the reaction forces at A and C for the 20 kg load.
Read the worked answer
Take moments about A, so that the unknown reaction at A has no moment. The beam weight acts at 2.5 m from A: M_beam = 25 × 9.81 × 2.5 = 613.1 N m clockwise. The 20 kg load acts at B, 6.0 m from A: M_load = 20 × 9.81 × 6.0 = 1177.2 N m clockwise. Total clockwise moment = 613.1 + 1177.2 = 1790.3 N m. The roller at C is 4.5 m from A and pushes upward, giving an anticlockwise moment 4.5R_C. Hence 4.5R_C = 1790.3, so R_C = 397.8 N.
Vertical equilibrium for the whole beam gives R_A + R_C = (25 + 20) × 9.81 = 441.45 N. Thus R_A = 441.45 − 397.8 = 43.6 N upward. Both reactions are positive, so the assumed upward directions are correct.
2. Determine the greatest load that can be hung at B before the beam is on the point of tipping about C.
Read the worked answer
Tipping about C occurs when the pivot at A is just about to lose contact, so R_A = 0 and the beam is held only by the roller at C. Take moments about C to eliminate R_C. The beam weight acts 4.5 − 2.5 = 2.0 m to the left of C: M_beam = 25 × 9.81 × 2.0 = 490.5 N m anticlockwise. Let the hanging load be W newtons at B, which is 6.0 − 4.5 = 1.5 m to the right of C: M_load = 1.5W clockwise. At the tipping threshold these moments balance: 1.5W = 490.5, so W = 327.0 N. The corresponding mass is m = 327.0 / 9.81 = 33.3 kg. So the greatest load is about 33 kg. Any larger load would create a net clockwise moment about C that the pivot at A could not counter by pushing upward.
3. If the hanging load is 40 kg, calculate R_A and R_C and interpret the sign of R_A.
Read the worked answer
Take moments about A with the 40 kg load at B: M_beam = 25 × 9.81 × 2.5 = 613.1 N m; M_load = 40 × 9.81 × 6.0 = 2354.4 N m; total clockwise moment = 2967.5 N m. Thus 4.5R_C = 2967.5, so R_C = 659.4 N. Vertical equilibrium gives R_A + R_C = (25 + 40) × 9.81 = 637.65 N, so R_A = 637.65 − 659.4 = −21.8 N. The negative sign means the pivot at A must pull downward on the beam. If the pivot can provide a downward force, equilibrium is possible; if it is only a simple upward support, contact is lost and the beam tips about C. This is consistent with the 33.3 kg tipping threshold calculated earlier, since 40 kg exceeds that value.