Provided is the question and previous work done for A, B, and C. All I need is D please. a)FBD of one cylinderForce equation in vertical direction:W = NFrictional force:F = μ_N = μ WForce equation in horizontal directionP+ F = MaWP =P = W= W ( ²1 – Ps)a FStep3c)b)Force equation in vertical direction:Force equation in vertical direction:W = NFrictional force:F = ₂N = kWForce equation in horizontal directionP-F = MaWP = a + F9= W ( ² + μk)P = WStep4d)c)Since the weight of cylinder and frictioncoefficient are not given, we have assumedsome valueMatlab code:close allclear all Since the weight of cylinder and frictioncoefficient are not given, we have assumedsome valueMatlab code:close allclear allclcW=10; %Enter weight of one cylinder in lbr=5; % Enter radius of cylinder in ftmu_s= 0.3; % Enter static friction coefficientmu_k=0.2; % Enter kinetic friction coefficienta= 10; % Enter acceleration of the cylinder inft/s2g=32.17; %Gravity in ft/s2I=0.5*(W/g)*r^2;nent of inertialF=(I*(a/r))/r; %Frictinal force using momentequationif(mu_s*W>F)P=W*((a/g)-mu_s);elseP=W*((a/g)+mu_k);endIdentical cylinders of weight and radiusare pushed by a series ofmoving arms. The coefficient of friction between all surfaces is. Denoting by the magnitude of the acceleration of the arms,derive an expression for (a) the maximum allowable value of ifeach cylinder is to roll without sliding, (b) the minimumallowable value of if each cylinder is to move to the rightwithout rotating.Part (c) Let, and between all surfaces. Determine the horizontalcomponent of the force exerted on each cylinder when theacceleration is (1) to the right and (2) 10 to the right.Benchmark solutions for part (c) are given as (1) 1.456 lb to theright, (2) 2.66 lb to the right. For part (c), your solution shouldbe coded into MATLAB so it can easily be run for alternatecases.Part (d) Plot the horizontal component of the force exerted oneach cylinder for different ranges of acceleration, whichillustrate the various case of rolling and sliding obtained in parts(a) and (b).

To Plot: The horizontal component of force for different ranges of acceleration. Given: The…

FBD of one cylinder N Force equation in vertical direction: W = N Frictional force: F = μ_N = μ W Force equation in horizontal direction P+ F = Ma P = -a-F P = W ( = −μ₂) Step3 c) b) Force equation in vertical direction: Force equation in vertical direction: W = N Frictional force: F = ₂N = ₂ W Force equation in horizontal direction. P-F = Ma P = Wa+F : W ( = + μK.) P=W Step4 d) c) Since the weight of cylinder and friction coefficient are not given, we have assumed some value Matlab code: close all

FBD of one cylinder N Force equation in vertical direction: W = N Frictional force: F = μ_N = μ W Force equation in horizontal direction P+ F = Ma P = -a-F P = W ( = −μ₂) Step3 c) b) Force equation in vertical direction: Force equation in vertical direction: W = N Frictional force: F = ₂N = ₂ W Force equation in horizontal direction. P-F = Ma P = Wa+F : W ( = + μK.) P=W Step4 d) c) Since the weight of cylinder and friction coefficient are not given, we have assumed some value Matlab code: close all

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter1: Introduction To Statics

Section: Chapter Questions

Problem 1.12P: A differential equation encountered in the vibration of beams is d4ydx4=2D where x = distance...

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