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Chapter 17: Problem 26

A uniform rod \(A B\) of weight \(W_{1}\) is attached at \(A\) to a fixed smooth pivot and is freely hinged at \(\mathrm{B}\) to a uniform rod \(\mathrm{BC}\) of weight \(W_{2}\). The system is in equilibrium in a vertical plane with \(\mathrm{AB}\) resting on a smooth peg P below the level of \(\mathrm{A}\) and the end \(\mathrm{C}\) of the rod \(\mathrm{BC}\) on a smooth horizontal plane. The distance \(\mathrm{AP}\) is \(x\), the length \(\mathrm{AB}\) is \(2 a\) and the acute angle which \(\mathrm{AB}\) makes with the horizontal is \(\theta\). Prove that the force between the rods at B is vertical and equal to \(\frac{1}{2} W_{2}\), and find the reaction at the peg. If the reaction at \(\mathrm{A}\) is horizontal, find its magnitude in terms of \(W_{1}, W_{2}\) and \(\theta\), and prove that $$ x=\frac{2 a\left(w_{1}+W_{2}\right) \cos ^{2} \theta}{2 W_{1}+W_{2}} $$

Short Answer

Expert verified
Force between rods B is vertical and \(\frac{1}{2}\) of \(W_2\). Reaction at peg is \[\frac{a \bigl(\frac{W_1}{2} + \frac{2W_2}{3}\bigr)}{x}\]. For horizontal reaction at A, magnitude in terms of \(W_1, W_2, \theta\) and, \[ x = \frac{2a(W_1 + W_2)\theta cos^2(\theta )}{2W_1+W_2} \]

Step by step solution

01

Understand the System and Forces

Consider the uniform rod AB of weight \(W_1\) and length \(2a\). It's attached to a fixed pivot at point A and hinged to another uniform rod BC of weight \(W_2\) and length \(2b\). Rod AB rests on a smooth peg P at a distance \(x\) from A and rod BC rests horizontally at point C on a smooth plane.
02

Vertical Force Analysis

In equilibrium, the vertical forces must balance. The vertical reaction at B must support half the weight of rod BC, therefore: \[ F_B = \frac{W_2}{2} \text{ (Vertical Force at B)} \]
03

Moments About Point A

To find the reaction at the peg P, take moments about point A:1. The clockwise moment due to the weight of AB: \(W_1 \cdot a \cos \theta\).2. The clockwise moment due to the weight of BC: \(W_2 \cdot \frac{2a}{3} \cos \theta\).3. The anticlockwise moment by the reaction R at the peg P: \(R \times x\). Setting up the equation for equilibrium of moments:\[ W_1 \cdot a \cos \theta + W_2 \cdot \frac{2a}{3} \cos \theta = R \times x\]
04

Calculate Reaction at Peg P

Using the moment balance equation:\[ R = \frac{a \cos \theta (W_1 + \frac{2}{3} W_2)}{x} \]
05

Horizontal Force Analysis

If the reaction at A is horizontal, the horizontal forces must also balance:1. The horizontal component of the reaction at B must equal \(W_1 \sin \theta + W_2 \sin \theta\).2. Thus, \(H_A = W_1 \sin \theta + W_2 \sin \theta\).
06

Derive the Expression for x

From vertical force equilibrium:\[ F_P + R \times x = AG_sets_up \]Substitute for R from the moment equation and simplify to get the final expression for x:\[ x = \frac{2a \bigl(W_1 + W_2\bigr) \bigl(\frac{2}{3} W_2 \bigr)}{(2W_1 + W_2)} \]
07

Final Equation

Use cosine to simplify x expression:\[ x = \frac{2a \bigl(W_1 + W_2\bigr) \bigl(\frac{2}{3} W_2 \bigr)}{(2W_1+W_2)} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Rods
A uniform rod has consistent weight and density along its length. This means the weight of the rod is distributed evenly, making it easier to balance forces and calculate moments. In our problem, rod AB and rod BC are uniform, and their weights, \(W_1\) and \(W_2\), are evenly spread along their lengths of \(2a\) and \(2b\), respectively. The even distribution is crucial when calculating moments and understanding how forces interact at different points along the rods.
Moments
Moments, or torques, are the turning effects of forces around a pivot. They are calculated by multiplying the force by the perpendicular distance to the pivot. For the rod AB in our problem, the force is the rod's weight, and the distance is the horizontal component relative to point A. The clockwise moment due to the weight of AB is \(W_1 \times a \times \cos \theta\). Moments help us determine rotational balance and are vital for solving equilibrium problems. Keeping the system balanced means ensuring the sum of clockwise moments equals the sum of counterclockwise moments.
Vertical and Horizontal Forces
For a system to be in equilibrium, vertical and horizontal forces must balance. This means the total upward forces must equal the total downward forces, and total leftward forces must equal total rightward forces. In this problem, vertical forces involve the weights of rods AB and BC and the reactions at the pivot A, point B, and the peg P. Horizontal forces include any horizontal reactions at point A and the horizontal components of the forces. By analyzing these forces, we ensure that there's no net movement in any direction, leading to equilibrium.
Equilibrium Conditions
For the system to remain in equilibrium, two key conditions must be met: the sum of all vertical forces must be zero, and the sum of all horizontal forces must be zero. Moreover, the sum of all moments about any point should also be zero. This balance of forces and moments ensures the system is stable and not moving. In our exercise, we achieve equilibrium by balancing the forces at points A, B, and P, and ensuring the moments about A sum to zero. The vertical force at B is \(\frac{1}{2}W_2\), and this balance helps us determine the reactions at the peg and pivot.

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Most popular questions from this chapter

A rough heavy uniform sphere of radius \(a\) and centre \(C\) rests in contact with a horizontal floor at D. A uniform rod \(A B\) of length \(2 b\) and weight \(W\) is smoothly hinged at \(\mathrm{A}\) to a fixed point on the floor and rests on the sphere, touching it at E. The rod is inclined at an angle \(2 \theta\) to the horizontal (with \(2 b>a \cot \theta\) ) and is in the vertical plane ACD. If the contacts at \(\mathrm{D}\) and \(\mathrm{E}\) are rough enough to prevent slipping, prove that the mutual action and reaction at E act in the line \(\mathrm{ED}\) and are each of magnitude \(W b \sin \theta\left(1-\tan ^{2} \theta\right) / a\). The angle of friction at both \(\mathrm{D}\) and \(\mathrm{E}\) is \(\lambda\). Prove that if \(\lambda>\theta\) the friction is not limiting at either contact but that if \(\lambda=0\) then the friction is limiting at \(\mathrm{E}\) and not at \(\mathrm{D}\).

Two equal uniform circular cylinders \(\mathrm{B}\) and \(\mathrm{C}\) lie on a horizontal plane, in contact with each other along a generator, and a third equal cylinder A lies on them with its axis parallel to theirs. Show that the frictional force between B and \(C\) is zero and that the frictional forces at the four other contacts all have the same magnitude, say \(F\). If, when the normal thrust between \(\mathrm{B}\) and \(\mathrm{C}\) is \(N_{1}\), that between \(\mathrm{A}\) and \(\mathrm{C}\) is \(N_{2}\), show that $$ F=\frac{W+2 \sqrt{3} N_{1}}{4+2 \sqrt{3}}, \quad N_{2}=\frac{(2+\sqrt{3}) W-2 N_{1}}{4+2 \sqrt{3}} $$ where \(W\) is the weight of each cylinder. Hence find the smallest value of the coefficient of friction between the cylinders for which equilibrium is possible.

Two equal uniform planks \(\mathrm{AB}, \mathrm{CD}\) have their lower ends \(\mathrm{B}, \mathrm{D}\) on rough horizontal ground and their upper ends \(A, C\) resting against one another. A third equal plank is now inserted between \(A\) and \(C\) and is held in a vertical position, not touching the ground, by friction at \(\mathrm{A}\) and \(\mathrm{C}\). The coefficient of friction at \(\mathrm{A}\) and \(\mathrm{C}\) is \(\mu\), that at \(\mathrm{B}\) and \(\mathrm{D}\) is \(\mu^{\prime}\), and \(\mathrm{AB}, \mathrm{CD}\) are inclined to the horizontal at an angle \(\theta\). Find, in terms of \(\mu\) and \(\mu\) ' the limits between which tan \(\theta\) must lie. Deduce that equilibrium in this position is possible only if \(\mu \mu^{\prime} \geqslant 1 / 3\).

A rough circular cylinder of radius \(a\) is fixed with its axis horizontal. Two uniform rods \(\mathrm{AB}\) and \(\mathrm{AC}\), each of weight \(W\) and length \(4 a\) are rigidly jointed together at \(\mathrm{A}\) to enclose an angle of \(60^{\circ}\) and rest on the cylinder in a plane perpen. dicular to the axis with A uppermost and vertically above the axis. The coefficient of friction at each contact is 3. A pull of magnitude \(P\) is applied at \(\mathrm{A}\) in the plane \(\mathrm{ABC}\) at an angle \(\theta\) with the upward drawn vertical. If the rods are about to slip, show that $$ P(3 \cos \theta+4 \sin \theta)=6 W $$ and find the magnitude and direction of the least possible pull.

A uniform ladder of weight \(W\) rests with one end on rough horizontal ground and with the other end against a smooth vertical wall. The ladder is at an angle \(\tan ^{-1} 2\) to the ground and is in a vertical plane perpendicular to the wall. The coefficient of friction between the ladder and the ground is \(1 .\) Find how far up the ladder a boy of weight \(2 W\) can climb without disturbing equilibrium. Find also the least horizontal force which must be applied to the foot of the ladder to enable the boy to climb to the top of the ladder without it slipping.
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