Question

1. What is the horizontal distance of the center of gravity of the system from the point where the ladder touches the ground?

2. What is the torque about the axis of rotation (point B) by taking the total weight of the person + ladder acting at the center of gravity?

Answer 1

1.The horizontal distance of the system's CG from point A is approximately 1.69 meters.

2. The torque about point B due to the combined weight of the person and ladder acting at the CG is 640 Nm.

Ladder length:= 4 m from the image).

Person's height = 1.8 m from the image).

Ladder angle = 71° from the image).

Applying the "weighted average" principle, considering the lengths of each section as weights:

CG_horizontal = (AC * CG_AC + CB * CG_CB) / (AC + CB)

CG_horizontal = (1.5 m * 0.5 m + 2.5 m * 2 m) / 4 m

CG_horizontal = 1.69 m

2.

Total weight (W) 800 N from the image and the ladder's weight

Lever arm (d) is the perpendicular distance from the CG to the line of action of W (which acts vertically through the CG).

From the image, d is the horizontal distance from the CG to point B = 0.8 m.

Then, the torque (τ) is:

τ = W * d

τ = (800 N + negligible) * 0.8 m

τ = 640 Nm