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Chapter 11: Problem 37

Two uniform planks, each of mass \(m\) and length \(L,\) are connected by a hinge at the top and by a chain of negligible mass attached at their centers, as shown in the figure. The assembly will stand upright, in the shape of an \(A,\) on a frictionless surface without collapsing. As a function of the length of the chain, find each of the following: a) the tension in the chain, b) the force on the hinge of each plank, and c) the force of the ground on each plank.

Short Answer

Expert verified
Based on the given problem and step-by-step solution, create a short answer question: Question: A teacher set up two identical planks on a hinge, and a single chain was attached to both planks with an equal length. Knowing the length of the chain, the mass of the planks, and the distance between the hinges, calculate the tension in the chain, the force on the hinge of each plank, and the force of the ground on each plank. Answer: To calculate the required forces, first identify all the forces acting on the planks and analyze them in the vertical and horizontal directions. Then, solve the equilibrium equations for forces perpendicular and parallel to the hinge to find the tension in the chain (T), the force on the hinge of each plank (F_H), and the force of the ground on each plank (F_v). The expressions for each force are as follows: a) Tension in the chain: \(T = \frac{mg}{\sqrt{1+\frac{L^2}{4d^2}}}\) b) Force on the hinge of each plank: \(F_H = \sqrt{F_h^2 + F_{h_v}^2}\) c) Force of the ground on each plank: \(F_v = F_h + T\frac{L}{2d}\)

Step by step solution

01

Step 1:Identify forces on the planks

We have the following forces acting on the planks: 1. The gravitational force acting at the center of the mass of each plank, \(mg\). 2. The tension in the chain acting at the center of each plank, \(T\). 3. The horizontal force exerted on each plank by the hinge, \(F_h\). 4. The vertical force exerted on each plank by the ground, \(F_v\). 5. The vertical force exerted on each plank by the hinge, \(F_{h_v}.\)
02

Examine forces acting perpendicular to the hinge

Let the angle between the plank and the vertical axis be denoted by \(\theta\). We'll analyze the forces acting perpendicular to the hinge point (\(F_v\) and the horizontal components of the tension and hinge force) and set up the equilibrium equation: \(\sum F_{perpendicular} = 0 = F_v - F_h - T\sin{\theta}\)
03

Examine forces acting parallel to the hinge

Now, we'll analyze forces acting parallel to the hinge point (\(F_{h_v}\) and the vertical components of the tension and gravitational force) and set up the equilibrium equation: \(\sum F_{parallel} = 0 = F_{h_v} - T\cos{\theta} + mg\)
04

Calculate the tension in the chain

Now, we will find the tension in the chain (\(T\)) using the vertical and horizontal components of the equilibrium equation: \(T\sin{\theta} = F_v - F_h\) \(T\cos{\theta} = mg - F_{h_v}\) Divide the two equations above: \(\tan{\theta} = \frac{F_v - F_h}{mg - F_{h_v}}\) We know that \(\tan{\theta}=\frac{L}{2d}\), where \(d\) is half the length of the chain. Therefore: \(T = \frac{mg}{cos{\theta}} = \frac{mg}{\sqrt{1+\frac{L^2}{4d^2}}}\)
05

Find the force on the hinge of each plank

From the equilibrium equations for forces perpendicular and parallel to the hinge: \(F_h = F_v - T\sin{\theta} = F_v - T\frac{L}{2d}\) \(F_{h_v} = mg - T\cos{\theta} = mg - T\frac{d}{\sqrt{L^2+4d^2}}\) We can find the magnitude of the total force on the hinge of each plank using the Pythagorean theorem: \(F_H = \sqrt{F_h^2 + F_{h_v}^2}\)
06

Find the force of the ground on each plank

From the equilibrium equation for forces perpendicular to the hinge: \(F_v = F_h + T\sin{\theta} = F_h + T\frac{L}{2d}\) So, we have found the tension in the chain, the force on the hinge of each plank, and the force of the ground on each plank: a) Tension in the chain: \(T = \frac{mg}{\sqrt{1+\frac{L^2}{4d^2}}}\) b) Force on the hinge of each plank: \(F_H = \sqrt{F_h^2 + F_{h_v}^2}\) c) Force of the ground on each plank: \(F_v = F_h + T\frac{L}{2d}\)

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