Problem 21 - An Introduction to the Mechanics of Solids in SI Units

Question

A rigid rod with negligible weight and small transverse dimensions carries a load \(W\) whose position is adjustable. The rod rests on a small roller at \(A\) and bears against the vertical wall at \(B\). Determine the distance \(x\) for any given value of \(\theta\) such that the rod will be in equilibrium. Assume that friction is negligible.

Step-by-Step Solution

Answer

The distance \(x\) for which the rod will be in equilibrium for any given value of \(\theta\) is \(x = (L\sin\theta)\cos\theta\).

Step 1: Identify forces and moments

There are three forces acting on the system: the force \(W\) acting downwards, the normal force \(N\) at point \(A\) acting upwards and the normal force at point \(B\) acting horizontally to the left. The moments are generated by the forces \(W\) and \(N\) about point \(A\).

Step 2: Apply conditions for equilibrium

For the rod to be in equilibrium the sum of the forces and the moments (torques) acting on it must be equal to zero. So, we need two equations: one for the sum of the forces and the other for the sum of the moments. The equation for the sum of forces is: \(N - W\cos\theta = 0\). The equation for the moments is: \(N(L\sin\theta) - Wx = 0\). Both equations contain \(N\) and \(W\), but we are supposed to find \(x\) for any given \(\theta\), hence we will eliminate \(N\) and \(W\) from these equations.

Step 3: Eliminate N and find x

By solving the equation for the sum of forces for \(N\): \(N = W\cos\theta\). Substituting \(N\) into the equation for the sum of moments gives us: \(W\cos\theta(L\sin\theta) - Wx = 0\) or, simplified, \((L\sin\theta)\cos\theta - x = 0\). Solving this equation for \(x\) yields the solution: \(x = (L\sin\theta)\cos\theta\).

Key Concepts

Equilibrium Conditions

In the study of statics, equilibrium conditions are fundamental. Equilibrium occurs when all forces and moments (also called torques) acting on an object result in zero net force and zero net moment. This means that an object at rest remains at rest, and an object in constant motion continues moving in a straight line without rotation.

For a rigid system like the rod in the exercise, two main conditions hold true for equilibrium:

• Sum of all forces in any direction must be zero.
• Sum of all moments about any point must also be zero.

These conditions allow us to solve for unknowns in the system, such as the distance of the weight on the rod. By using the equilibrium equations, we can find values that satisfy both the forces and moments keeping the system stable and unmoving.

Force Analysis

Force analysis involves identifying and calculating all the forces acting on a system. In this exercise, we can break it down step by step:

First, identify the forces involved:

• The weight force, \( W \), which acts vertically downwards.
• The normal force at point \( A \), \( N \), which acts vertically upwards.
• The horizontal normal force at point \( B \), which pushes the rod against the wall.

These forces interact and must be analyzed together to maintain the rod's equilibrium condition. By calculating each of these forces and ensuring they sum to zero, we prevent any non-allowed movement in the system.

For example, the vertical forces must balance out, meaning the upward \( N \) must exactly counteract the component of \( W \) acting vertically, leading to the equation \( N - W\cos\theta = 0 \). This ensures the rod doesn't move vertically when forces are in balance.

Moment Calculation

Moment calculation is crucial when analyzing rotational equilibrium. A moment, or torque, is the measure of a force causing an object to rotate about a point. In our exercise, the rod can pivot at point \( A \), so we calculate moments about this point.

For rotation to be in equilibrium, the sum of moments around the pivot point must be zero. Let's consider:

• The moment caused by \( N \): It tries to rotate the rod due to the force applied at a certain distance from point \( A \). This distance is considered as \( L\sin\theta \).
• The moment caused by \( W \): It also creates a torque around point \( A \), acting at a lever arm distance of \( x \).

To maintain equilibrium, these torques should balance out, leading us to the equation \( N(L\sin\theta) - Wx = 0 \). Simplifying this helps solve for \( x \) in terms of \( L \) and \( \theta \), determining the position of \( W \) that keeps the rod still and balanced.