Question

A system of parallel forces are acting on a rigid bar as shown in fig. Reduce this system to

: (i) a single force, (ii) a single force and a couple at A(iii) a single force and a couple at B.​

Answer 1

Answer:

(i) A single force: A single force means just to find out the resultant of the system. Since there are parallel force i.e resultant is sum of vertical forces, R = 32.5 –150 + 67.5 –10 = -60 R = (∑V2)1/2 R = 60N Let d = Distance of resultant from A towards right. To find out location of resultant apply varignon’s theorem : R.d = 150 × 1 – 67.5 × 2 + 10 × 3.5 60.d = 150 × 1 – 67.5 × 2 + 10 × 3.5 d = 0.833 m i.e resultant is at a distance of 0.83 m from A. (ii) A single force and a couple at A: It means the whole system is to convert in to a single force and a single couple. Since we convert all forces in to a single force i.e. resultant. Now apply two equal and opposite force i.e. ‘R’ at point A. Now force ‘R’ which is act at point E and upward force which is act at point A makes a couple of magnitude, Magnitude = Force × distance = 60 × 0.833 = 49.98 Nm and a single force of magnitude = 60N (iii) A single force and a couple at B: Since AE = 0.833 m, then BE = 3.5 – 0.833 = 2.67 m Now, the force R = –60N is moved to the point B, by a single force R = –60N and a couple of magnitude = R × BE = –60 × 2.67 = 160 Nm. Hence single force is 60 N and couple is 160 NmRead more on Sarthaks.com - https://www.sarthaks.com/508299/system-parallel-forces-acting-rigid-bar-as-shown-in-fig-reduce-this-system-to-single-force