Physics: Fence Post Tension and Normal Force Calculation

Question:

A fence post of mass m = 6 kg supports a fence with three lengths of barbed wire. The bottom wire is a distance d = 0.35 m from the ground, and each wire is spaced 0.35 m above the previous one. Each wire exerts the same horizontal tension T on the post.

An additional guy wire is used to keep the post upright, applying a force of F_a = 450 N. The guy wire is attached at a height h = 0.86 m from the ground and makes an angle θ = 45° with the horizontal.

(1) Write an expression for the tension T in any one of the three fence wires.
(2) Calculate the tension in Newtons.
(3) What is the normal force that the ground exerts on the post?

Answer:

Step 1: Concept of Torque and Force Balance

We use the principle of static equilibrium. For the post to remain stationary:

Net torque about the base = 0
Net vertical force = 0

Step 2: Writing the Torque Equation (Counterclockwise Positive)

Στ = 0 = -dT – 2dT – 3dT + h × F_a × cos(θ)
⇒ -6dT + h × F_a × cos(θ) = 0

Solving for T:

T = (h × F_a × cos(θ)) / (6 × d)

Step 3: Substituting Values

h = 0.86 m
F_a = 450 N
θ = 45° → cos(45°) = √2 / 2 ≈ 0.7071
d = 0.35 m

T = (0.86 × 450 × cos(45°)) / (6 × 0.35)
T = (387.81) / 2.1 ≈ 184.7 N

✅ Tension in one fence wire ≈ 184.7 N

Step 4: Calculating the Normal Force

Apply vertical force balance:

ΣF_vertical = 0 ⇒ F_n = mg + F_a × sin(θ)

m = 6 kg
g = 9.81 m/s²
F_a = 450 N
sin(45°) = 0.7071

F_n = (6 × 9.81) + (450 × 0.7071)
F_n ≈ 58.86 + 318.2 = 377.06 N

✅ Normal Force from the ground ≈ 377.06 N

Final Answers:

Expression for Tension: T = (h × F_a × cos(θ)) / (6 × d)
Tension in a wire: 184.7 N
Normal force: 377.06 N

Conclusion:

By analyzing torque and vertical force equilibrium, we calculated the tension in each of the fence wires as well as the total normal force exerted by the ground. The guy wire provides stability by balancing the torque caused by horizontal wire tensions.

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