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Free body diagram; reactions on two cylinders on incline

Short answer

A free body diagram isolates a body showing all external forces. Assuming cylinder B diameter 90 mm, geometry gives reaction angles. Equilibrium yields: wall reaction on A 139.9 N horizontal, incline on A 167.0 N, incline on B 164.0 N, contact force between cylinders 86.3 N.

What is free body diagram? The cylinders A and B rest on an inclined surface which makes an angle of 25° with horizontal as shown in figure below. Determine reaction at contact points. Weight of cylinder A (WA) = 100 N, Weight of cylinder B (WB) = 200 N, Diameter of cylinder A (rA) = 60 mm, Diameter of cylinder B (rB) = 9 (as printed, unit cut off). [Figure: cylinder A against a vertical wall and the 25° incline, cylinder B resting on A and on the incline.]

Free body diagram

A free body diagram is a sketch of a body separated from its surroundings, showing all external forces and moments acting on it: weight, applied loads and all reactions at supports and contacts.

Data and assumptions

Incline angle 25° with the horizontal. W_A = 100 N, W_B = 200 N. Cylinder A (size 60 mm) touches the vertical wall and the incline. Cylinder B rests on the incline and on A. All surfaces are smooth. The diameter of B is printed as "9…" (cut off). I assume d_B = 90 mm, which fits the figure, where B is larger than A. (Sizes enter only through the angle of the line of centres.)

Geometry

Take the corner where the wall meets the incline as origin. Direction along the incline u = (cos 25°, sin 25°), normal to the incline n = (−sin 25°, cos 25°).

  • Centre of A: x = r_A = 30 mm, and distance r_A from the incline, so y_A = r_A(1 + sin 25°)/cos 25° = 47.09 mm. So C_A = (30, 47.09).
  • Centre of B lies at distance r_B = 45 mm from the incline and 75 mm (r_A + r_B) from C_A. Solving gives C_B = (90.26, 91.74).
  • The line of centres B to A has direction (−0.803, −0.595), that is 36.5° below the horizontal.

Cylinder B (weight 200 N, incline reaction R_B along n, contact force F from A)

F acts on B along the line of centres, away from A: direction (0.803, 0.595).

x:  −R_B sin25° + 0.803 F = 0
y:   R_B cos25° + 0.595 F − 200 = 0

Solving: R_B = 164.0 N, F = 86.3 N.

Cylinder A (weight 100 N; wall reaction R_w horizontal; incline reaction R_A along n; F from B)

F acts on A along (−0.803, −0.595).

y:  R_A cos25° − 0.595 × 86.27 − 100 = 0  ->  R_A = 167.0 N
x:  R_w − R_A sin25° − 0.803 × 86.27 = 0  ->  R_w = 139.9 N

Check on the whole system: the vertical components of the incline reactions = (167.0 + 164.0) cos 25° = 300 N = 100 + 200 ✓. Horizontal: (167.0 + 164.0) sin 25° = 139.9 N = R_w ✓.

Answer (with d_B = 90 mm): wall on A = 139.9 N (horizontal); incline on A = 167.0 N; incline on B = 164.0 N (both perpendicular to the incline); contact force between A and B = 86.3 N.

✓ Answer checked by Claude Opus 5.5 (high effort)

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