A Post Is Supported By Two Wires

A post is supported by two wires is a classic static‑equilibrium problem that appears in introductory physics and engineering courses. The scenario typically involves a vertical pole or post that is held upright by two cables (or wires) attached to the top of the post and anchored to the ground at different points. Because the post is not moving, the sum of all forces and the sum of all moments acting on it must be zero. By resolving the tension forces in each wire into their horizontal and vertical components, we can determine the unknown tensions, the angle each wire makes with the post, or the required length of the wires for a given load. Understanding this setup builds a foundation for analyzing more complex structures such as transmission towers, crane booms, and suspension bridges.

1. Core Concepts Behind the Problem When a post is supported by two wires, the system is in static equilibrium. This means:

• Translational equilibrium: The vector sum of all forces equals zero.
  [ \sum \vec{F}=0 \quad\Rightarrow\quad \sum F_x=0,;\sum F_y=0,;\sum F_z=0 ]
• Rotational equilibrium: The sum of all moments (torques) about any point equals zero.
  [ \sum \vec{M}=0 ]

In most textbook versions, the post is assumed to be weightless or its weight is given as a known downward force acting at its centre of gravity. The wires are considered massless, inextensible, and perfectly flexible, so they can only transmit tension along their length. No compression or bending stiffness is assumed in the wires themselves.

Key variables that often appear:

2. Setting Up the Free‑Body Diagram

The first step in solving “a post is supported by two wires” is to draw a clear free‑body diagram (FBD):

1. Represent the post as a vertical line.
2. Indicate the weight (W) acting downward at the post’s centre (or at the top if a point load is applied).
3. Draw the two wires as straight lines emanating from the attachment point on the post to their ground anchors.
4. Label the tension forces (T_1) and (T_2) along each wire, pointing away from the post (since wires pull).
5. Show the angles (\theta_1) and (\theta_2) measured from the vertical (or from the horizontal, depending on convenience).
6. Include the reaction forces at the base of the post if the post is not fixed; often the base is assumed to be a pin or a smooth surface, giving a horizontal reaction (R_x) and a vertical reaction (R_y). In many simplified problems the base is fixed, so the reactions are taken care of by the moment equilibrium.

3. Writing the Equilibrium Equations ### 3.1 Force Balance

Resolve each tension into vertical and horizontal components:

• Wire 1:
  [ T_{1x}=T_1\sin\theta_1 \quad (\text{horizontal})\ T_{1y}=T_1\cos\theta_1 \quad (\text{vertical}) ]
• Wire 2:
  [ T_{2x}=T_2\sin\theta_2\ T_{2y}=T_2\cos\theta_2 ]

Assuming the base of the post is fixed (no translation), the horizontal forces must cancel:

[ \sum F_x = T_1\sin\theta_1 - T_2\sin\theta_2 = 0 \quad\Rightarrow\quad T_1\sin\theta_1 = T_2\sin\theta_2 \tag{1} ]

The vertical forces must balance the weight (W) (downward) and any vertical reaction at the base (R_y):

[ \sum F_y = T_1\cos\theta_1 + T_2\cos\theta_2 - W + R_y = 0 ]

If the base is a pin that can provide a vertical reaction, we often solve for (R_y) after finding the tensions. For a weightless post with only an external load (W) applied at the top, we set (R_y=0) and obtain:

[ T_1\cos\theta_1 + T_2\cos\theta_2 = W \tag{2} ]

3.2 Moment Balance

Taking moments about the base of the post eliminates the unknown reactions at that point. The moment contributed by a force is the force magnitude times its perpendicular distance to the pivot.

• The weight (W) creates a clockwise moment: (M_W = W \times \frac{L}{2}) (if the weight acts at the midpoint) or (W \times L) if it acts at the top. * Each wire’s tension creates a counter‑clockwise moment because its line of action passes above the base. The perpendicular distance from the base to the line of action of (T_i) is the horizontal offset (d_i). Thus the moment from wire i is (M_{T_i}=T_i \times d_i).

Setting the sum of moments to zero (counter‑clockwise positive):

[ \sum M_{base}= T_1 d_1 + T_2 d_2 - W \times \text{(lever arm of }W) = 0 \tag{3} ]

If the weight acts at the top, the lever arm is simply (L). If it acts at the centre, use (L/2).

Equations (1)–(3) provide three independent relationships for the three unknowns (T_1), (T_2), and either (R_y) or an angle, depending on what is given.

4. Step‑by‑Step Solution Procedure

Below is a generic algorithm you can follow whenever you encounter “a post is supported by two wires”:

1. Draw the FBD and label all known quantities (weights, angles, distances).
2. Choose a coordinate system (usually x‑horizontal, y‑vertical).
3. Write the force‑balance equations ((\sum F_x=0), (\sum F_y=0)). 4. Write the moment‑balance equation about a convenient point (often the base).
4. Solve the linear system for the unknown tensions.
5. Check the solution by substituting back into the original equations or by verifying that all forces and moments indeed sum to zero.
6. Interpret the results: Are the tensions realistic? Do any exceed the wire’s tensile strength? Adjust angles or lengths if necessary.

5. Worked Numerical Example

Let’s apply the procedure to a concrete problem:

*A 4