Questions: The figure shows a 16.9-kg, uniform ladder of length L hinged to a horizontal platform at point P1 and anchored with a steel cable attached at the ladder's midpoint. The angle alpha between the ladder and the floor is 50.0 degrees, and the angle theta between the cable and the floor is 300 degrees. Calculate the tension in the cable when a 80.9-kg person is standing three-quarters of the way up the ladder. N

Solution Steps

Here are the forces acting on the ladder:

• W_l: Weight of the ladder, acting downwards at the center of the ladder (L/2).
• W_p: Weight of the person, acting downwards at a distance of (3/4)L from the bottom of the ladder.
• T: Tension in the cable, acting at an angle θ = 30° with the horizontal, at a distance L/2 from the bottom of the ladder.
• F_h: Horizontal force from the hinge, acting at point P_1.
• F_v: Vertical force from the hinge, acting at point P_1.

Since the ladder is in equilibrium, the sum of the torques around P_1 is zero.

Consider the torques about the hinge point P_1. Taking counterclockwise torques as positive, the equation becomes:

T * (L/2) * sin(α - θ) - W_l * (L/2) * cos(α) - W_p * (3L/4) * cos(α) = 0

Where:

• α = 50°
• θ = 30°
• W_l = m_l * g = 16.9 kg * 9.8 m/s² = 165.62 N
• W_p = m_p * g = 80.9 kg * 9.8 m/s² = 792.82 N

Substitute the given values into the torque equation:

T * (L/2) * sin(20°) - 165.62 N * (L/2) * cos(50°) - 792.82 N * (3L/4) * cos(50°) = 0

Notice L/2 can be factored out from each term, and therefore cancelled from the equation.

T * sin(20°) - 165.62 N * cos(50°) - 792.82 N * (3/2) * cos(50°) = 0

T * sin(20°) = 53.24 N + 765.82 N = 819.06 N T = 819.06 N / sin(20°) T ≈ 2391.75 N

Final Answer: 2391.75 N