A Ladder of length L=4.75 m leans against a frame house at an angle 0 = 70 degrees. Frictionless contactsoccurs with a wall of height h = 4.3 m. The ladder has a mass m₁ = 16kg. The service worker who must climbthe ladder has a mass of m₂ = 60 kg Assume the contact point with the ground has a coefficient of frictionbetween the ladder footing and the ground of μ = 0.37.Compute the friction force needed to keep the ladder in equilibrium when the worker is at the top of theladder. What is the maximum friction that can be supported in this configuration? Note: Express answers asmagnitudes only and ignore signs/directions.FequilibriumTTH140.84140.84LDOTN 2x0%

Introduction In this problem, we will calculate the friction force for the equilibrium of the ladder when the worker is on top of it, as well as the maximum friction force that the given configuration can support, for which we will first draw the ladder's free body diagram and then calculate the friction force using the equilibrium equations. Given data: length of ladderL=4.75 minclination angleθ=70°height of the wallh=4.3 mmass of ladderm1=16 kgmass of workerm2=60 kgcoefficient of frictionμ=0.37 Find out: The friction force is needed to keep the ladder in equilibrium when the worker is on the top Fequilibrium The maximum friction force that can be supported in the given configuration FmaxCalculation of the friction force is needed to keep the ladder in equilibrium when the worker is on the top (F)equilibrium For the calculation of equilibrium force the following things we have to consider. The free-body diagram of the ladder with the wall is shown below, Taking moment about point B. ∑MA=0Wworker×4.75×cos70+wladder×0.8123-R×4.75×sin70=060 kg×9.81 m/s2×4.75×cos70+16 kg×9.81 m/s2×0.8123=R×4.75×sin70956.237 N+127.50 N=R×4.75×sin70R=242.798 N The friction force for equilibrium (F)equilibrium By considering the net force in the horizontal direction. ∑Fx=0R=μ×Nμ×N=Fequilibrium=equlibrium friction forceFequilibrium=R(F)equilibrium=242.798 N Hence the friction force needed to keep the ladder in equilibrium is 242.798 N.Calculation of the maximum friction force that can be supported in the given configuration (F)max By considering the net force in the vertical direction. ∑Fy=0N-Wworker-wladder=0N-60kg×9.81 m/s2-16kg×9.81 m/s2=0N=745.56 NFmax=maximum friction forceFmax=μ×NFmax=0.37×745.56 N(F)max=275.857 N Hence the maximum friction force that can supported is 275.857 N. Results: The friction force is needed to keep the ladder in equilibrium when the worker is on the top (F)equilibrium=242.798 N The maximum friction force that can be supported in the given configuration (F)max=275.857 N

Engineering

Mechanical Engineering

A Ladder of length L=4.75 m leans against a frame house at an angle 0 = 70 degrees. Frictionless contacts occurs with a wall of height h = 4.3 m. The ladder has a mass m₁ 16kg. The service worker who must climb the ladder has a mass of m₂ = 60 kg Assume the contact point with the ground has a coefficient of friction between the ladder footing and the ground of μ8 = 0.37. Compute the friction force needed to keep the ladder in equilibrium when the worker is at the top of the ladder. What is the maximum friction that can be supported in this configuration? Note: Express answers as magnitudes only and ignore signs/directions. μ Fequilibrium 140.84 Fmaz 140.84 L 0 = N N ? *0% T h x 0%

A Ladder of length L=4.75 m leans against a frame house at an angle 0 = 70 degrees. Frictionless contacts occurs with a wall of height h = 4.3 m. The ladder has a mass m₁ 16kg. The service worker who must climb the ladder has a mass of m₂ = 60 kg Assume the contact point with the ground has a coefficient of friction between the ladder footing and the ground of μ8 = 0.37. Compute the friction force needed to keep the ladder in equilibrium when the worker is at the top of the ladder. What is the maximum friction that can be supported in this configuration? Note: Express answers as magnitudes only and ignore signs/directions. μ Fequilibrium 140.84 Fmaz 140.84 L 0 = N N ? *0% T h x 0%

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Author:Sadiku, Matthew N. O.

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Basic Terminology in Mechanics

The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its environment. This branch of physics keeps its roots in Ancient Greece in the form of the scripts of Aristotle and Archimedes.

Question

A Ladder of length L=4.75 m leans against a frame house at an angle 0 = 70 degrees. Frictionless contacts
occurs with a wall of height h = 4.3 m. The ladder has a mass m₁ = 16kg. The service worker who must climb
the ladder has a mass of m₂ = 60 kg Assume the contact point with the ground has a coefficient of friction
between the ladder footing and the ground of μ = 0.37.
Compute the friction force needed to keep the ladder in equilibrium when the worker is at the top of the
ladder. What is the maximum friction that can be supported in this configuration? Note: Express answers as
magnitudes only and ignore signs/directions.
Fequilibrium
TTH
140.84
140.84
L
DO
T
N 2x0%

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