Two spheres A and B weight 1000 N and 750 N. Determine the reaction at all contact points 1, 2, 3 and 4. \(r_A = 400\) mm, \(r_B = 300\) mm.[Figure: a container with a vertical wall on the left, a horizontal floor, and a right wall inclined at 60° to…

Assumption (from the sketch): the lower sphere is B (r = 300 mm, 750 N) and the upper sphere is A (r = 400 mm, 1000 N). Contact points: 1 = B with left wall, 2 = B with floor, 3 = A with B, 4 = A with inclined wall.

Geometry: centre of B is 300 mm from the wall; centre of A is 700 mm from the wall. Horizontal distance between centres = 700 − 300 = 400 mm; distance between centres = 400 + 300 = 700 mm.

\(\cos\alpha = 400/700 \Rightarrow \alpha = 55.15^\circ\) (line of centres with horizontal).

FBD of A: W = 1000 N ↓; \(R_3\) along the line of centres from B towards A (55.15° above horizontal, to the right); \(R_4\) normal to the 60° wall, i.e. 30° above the horizontal pointing to the left.

\(\sum F_x = 0:\ R_3\cos55.15^\circ = R_4\cos30^\circ \Rightarrow R_4 = 0.6598R_3\)

\(\sum F_y = 0:\ R_3\sin55.15^\circ + R_4\sin30^\circ = 1000 \Rightarrow 0.8207R_3 + 0.3299R_3 = 1000\)

\(R_3 = 869.1\) N, \(R_4 = 573.5\) N

FBD of B: W = 750 N ↓, \(R_3 = 869.1\) N acting on B down-left at 55.15°, \(R_1\) horizontal (→) from the wall, \(R_2\) vertical (↑) from the floor.

\(R_1 = R_3\cos55.15^\circ = 869.1 \times 0.5714 = 496.6\) N

\(R_2 = 750 + R_3\sin55.15^\circ = 750 + 713.3 = 1463.3\) N

Check (whole system): \(\sum F_x: R_1 - R_4\cos30^\circ = 496.6 - 496.6 = 0\); \(\sum F_y: R_2 + R_4\sin30^\circ = 1463.3 + 286.7 = 1750 = 1000 + 750\). ✔

Answer: \(R_1 = 496.6\) N, \(R_2 = 1463.3\) N, \(R_3 = 869.1\) N, \(R_4 = 573.5\) N.