Question

Consider a horizontal triangular table with each vertex angle less than \(120^{\circ}\). At the vertices are frictionless pulleys over which pass strings knotted at \(P\), each with a weight \(W\) attached as shown in Figure \(20 .\) Show that at equilibrium the three angles at \(P\) are equal; that is, show that \(\alpha+\beta=\alpha+\gamma=\beta+\gamma=120^{\circ}\).

Short answer

In equilibrium, the angles at \( P \) are equal: \( \alpha + \beta = \alpha + \gamma = \beta + \gamma = 120^{\circ} \).

Step-by-step solution

Understand the Problem

We have a triangular table with weights tied at a point via strings over frictionless pulleys at each vertex. We need to prove that at equilibrium, the angles ( \( \alpha, \beta, \gamma \) ) at point \( P \) are such that each pair sums to \( 120^{\circ} \).

Identify Equilibrium Conditions

At equilibrium, the net force at point \( P \) must be zero. This implies that the vector sum of tensions in the strings must equilibrate. Each tension pulls away at angle \( \alpha, \beta, \gamma \) respectively.

Analyze Vector Forces

Consider the forces (tensions) \( T_1, T_2, T_3 \) acting along the directions of \( \alpha, \beta, \gamma \). Since the setup is symmetric and the pulleys are frictionless, the tension force in each string is the same magnitude as the weight \( W \), hence \( T_1 = T_2 = T_3 = W \).

Apply Geometry and Trigonometry

For equilibrium in the plane, the sum of the moments or angles due to these forces should satisfy the symmetry condition. From physics and construction symmetry of triangle plus the requirement \( \sin(\alpha) = \sin(\beta) = \sin(\gamma) \), gives the condition that \( \alpha = \beta = \gamma \).

Derive and Conclude Equal Angles

Given the triangle at \( P \) is equilateral in context of force directions because \( \alpha = \beta = \gamma \). Thus \( \alpha + \beta = \beta + \gamma = \gamma + \alpha = 120^{\circ} \), proving the required conditions for equilibrium in terms of equal angles.

Key concepts

Vector Sum of Forces

In any system, when we talk about equilibrium, one of the primary ideas is the vector sum of forces. A force is a vector since it has both magnitude and direction. At the point of equilibrium, the vector sum of all these forces equals zero. This means all forces acting in and out of the system need to balance perfectly.

In our triangular table setup, three forces come into play due to the weights attached at point \( P \). Each force represents the tension in the strings pulling outwards. These forces must counterbalance each other to ensure equilibrium. The concept here is when added together as vectors, the resultant of these forces should be zero, ensuring that the system doesn't move. Thus, understanding the nature and direction of these forces is crucial for determining how and when equilibrium occurs.

Frictionless Pulleys

A crucial aspect of this exercise is the use of frictionless pulleys. Frictionless pulleys are idealized devices where the pulleys exert no frictional forces on the strings that pass over them. As a result, the tension in the string remains constant on both sides of the pulley.

This consistency in tension is significant in our triangular table problem, as it ensures that the weight \( W \) suspended at different vertices is transmitted uniformly through the strings. Without friction in the pulleys, the forces can be perfectly communicated across the system. This assumption of frictionlessness helps simplify the mathematical model, letting us focus solely on the geometry and forces without worrying about any energy loss due to friction.

Angles in a Triangle

Let's delve into the geometric aspect of angles in a triangle. In geometry, the sum of angles inside a triangle is always \(180^{\circ}\). However, in our case, we are concerned with angles at the central knot \( P \). These angles determine how the forces spread out from \( P \).

In the given problem scenario, you are asked to demonstrate through equilibrium conditions that the angles \( \alpha, \beta, \) and \( \gamma \) at point \( P \) should each sum to \(120^{\circ}\) when paired. This specific configuration occurs because the angled tensions create a symmetrical force distribution, resulting in equal angles when viewed in terms of force balance. Such a geometric insight is crucial for understanding how equilibrium is achieved through symmetry.

Equilibrium Conditions

Equilibrium is the state where no net force acts on the system, resulting in no movement. In a static scenario, equilibrium conditions denote that all forces and moments must balance. It is the state where the system remains constant.

In our triangle table problem, equilibrium is shown by making sure that the vector sum of the forces is zero. This asks for symmetrical angles at the knot \( P \), leading each pair of angles to add up to \(120^{\circ}\). The conditions of equilibrium are achieved here by focusing on force distribution and angle equality, given the identical tensions because of frictionless pulleys and the weights attached. Properly understanding these conditions can help analyze various static systems effectively.