A horizontal rod with a mass of 10 kg and length 12 m is hinged to a wall at one end and supported by a cable which makes an angle of 30º with the rod at its other end. Calculate the tension in the cable and the force exerted by the hinge.

Introduction

This is a problem involving static equilibrium. The horizontal rod is supported at one end by a hinge and at the other end by a cable inclined at an angle. To solve this, we apply the conditions of equilibrium:

Sum of all horizontal forces = 0
Sum of all vertical forces = 0
Sum of all torques (moments) = 0

Given Data

Mass of rod, m = 10 kg
Length of rod, L = 12 m
Angle between cable and rod = 30º
Acceleration due to gravity, g = 9.8 m/s²
Weight of rod, W = mg = 10 × 9.8 = 98 N

Step 1: Torque Equation (About the Hinge)

Let T be the tension in the cable. The weight of the rod acts at its center (L/2 = 6 m from hinge). The tension provides a vertical component (Tsinθ) which produces torque in the opposite direction to the weight.

∑τ = 0 ⇒ T × L × sin(30º) = W × (L/2)

T × 12 × 0.5 = 98 × 6

T × 6 = 588

T = 588 / 6 = 98 N

Step 2: Vertical Force Balance

Let the vertical reaction at the hinge be R_v.

∑F_vertical = 0 ⇒ R_v + Tsin(30º) = W

R_v + 98 × 0.5 = 98

R_v + 49 = 98 ⇒ R_v = 49 N

Step 3: Horizontal Force Balance

Let the horizontal reaction at the hinge be R_h.

∑F_horizontal = 0 ⇒ R_h = Tcos(30º)

R_h = 98 × cos(30º) = 98 × 0.866 ≈ 84.87 N

Step 4: Total Reaction Force at the Hinge

R = √(R_h² + R_v²) = √(84.87² + 49²) ≈ √(7201.6 + 2401) ≈ √9602.6 ≈ 98 N

Final Answers

Tension in the cable: 98 N
Horizontal force by hinge: 84.87 N
Vertical force by hinge: 49 N
Total hinge force: ~98 N

Conclusion

By applying the principles of static equilibrium, we determined that the tension in the supporting cable is 98 N and the hinge provides both horizontal and vertical support to maintain the system in balance. This problem illustrates the practical application of torque and force balance in real-world structures.