Only need A, B, and C please. Identical cylinders of weight and radiusare pushed by a series ofmoving arms. The coefficient of friction between all surfaces is. (Updated to add: you may differentiate between and ifnecessary). Denoting by the magnitude of the acceleration ofthe arms, derive an expression for (a) the maximum allowablevalue of if each cylinder is to roll without sliding, (b) theminimum allowable value of if each cylinder is to move to theright without rotating.Part (c) Let, and between all surfaces (updated to add: you mayassume if necessary). Determine the horizontal component ofthe force exerted on each cylinder when the acceleration is (1) tothe right and (2) 10 to the right. Benchmark solutions for part(c) are given as (1) 1.456 lb to the right, (2) 2.66 lb to the right.For part (c), your solution should be coded into MATLAB so itcan easily be run for alternate cases.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). SAMA cord is wrapped around the inner drum of a wheel and pulled hori-zontally with a force of 40 lb. The wheel weighs 100 lb and has a radiusof gyration of 3.5 in. Knowing that μ, = 0.20 and = 0.15, determinethe acceleration of G and the angular acceleration of the wheel.a. Assume Rolling without Sliding. In this case, we haveà = rα = ₂aBy comparing the friction force obtained with the maximum frictionforce, we shall determine whether this assumption is justified. The massand moment of inertia of the wheel are111 =- ΣF, = md:Equations of Motion+) EMc = Σ(Me)ett:10032.2I = mk²649= 3.11 lb-sec²/ft+) EM, = la:=100 3.532.2 12(40)(2) ma() + Ia6.67= (3.11)(a)()+ 0.264aa +8.30 radians/sec²à ra = () (8.30)= 3.46 ft/sec²māF + 40F+40F= 29.3 lbN100 0+1EF, = 0:Maximum Available Friction Force= 0.264 lb-ft-sec²(3.11)(3.46)Fmax = μ,N= (0.20) (100) = 20 lbSince F> Fmax, the assumed motion is impossible.b. Rolling and Sliding. Since the wheel must roll and slide atthe same time, we draw a new diagram, where a and a are inde-pendent and whereF = 29.3 lb-N = 100 lb ↑F = F₁ = ₂N = (0.15) (100) = 15 lbFrom the computation of part a, it appears that F should be di-rected to the left.42F, = må:40 15 3.11aà +8.04 ft/sec²à 8.04 ft/sec²(40) () = 0.264a(15))a = 14.2 radians/sec² a = 14.2 radians/sec²100 lbF= 15 lbGIN100 lbOGс5 in.IN=40 lb2 in.3 in.r= 5 in.C40 lb5.0₂40 lb12 in.āmā5 in.ma5 in.

Force required Matlab codec) Since the weight of cylinder and friction coefficient are not given, we have assumed some value Matlab code: close all clear all clc W=10; %Enter weight of one cylinder in lb r=5; %Enter radius of cylinder in ft mu_s= 0.3; %Enter static friction coefficient mu_k=0.2; %Enter kinetic friction coefficient a= 10; %Enter acceleration of the cylinder in ft/s2 g=32.17; %Gravity in ft/s2 I=0.5*(W/g)*r^2; %Moment of inertial F=(I*(a/r))/r; %Frictinal force using moment equation if(mu_s*W>F) P=W*((a/g)-mu_s); else P=W*((a/g)+mu_k); end

Engineering

Mechanical Engineering

Identical cylinders of weight and radiusare pushed by a series of moving arms. The coefficient of friction between all surfaces is . (Updated to add: you may differentiate between and if necessary). Denoting by the magnitude of the acceleration of the arms, derive an expression for (a) the maximum allowable value of if each cylinder is to roll without sliding, (b) the minimum allowable value of if each cylinder is to move to the right without rotating. Part (c) Let, and between all surfaces (updated to add: you may assume if necessary). Determine the horizontal component of the force exerted on each cylinder when the acceleration is (1) to the right and (2) 10 to the right. Benchmark solutions for part (c) are given as (1) 1.456 lb to the right, (2) 2.66 lb to the right. For part (c), your solution should be coded into MATLAB so it can easily be run for alternate cases. Part (d) Plot the horizontal component of the force exerted on each cylinder for different ranges of acceleration, which illustrate the various case of rolling and sliding obtained in parts (a) and (b).

Identical cylinders of weight and radiusare pushed by a series of moving arms. The coefficient of friction between all surfaces is . (Updated to add: you may differentiate between and if necessary). Denoting by the magnitude of the acceleration of the arms, derive an expression for (a) the maximum allowable value of if each cylinder is to roll without sliding, (b) the minimum allowable value of if each cylinder is to move to the right without rotating. Part (c) Let, and between all surfaces (updated to add: you may assume if necessary). Determine the horizontal component of the force exerted on each cylinder when the acceleration is (1) to the right and (2) 10 to the right. Benchmark solutions for part (c) are given as (1) 1.456 lb to the right, (2) 2.66 lb to the right. For part (c), your solution should be coded into MATLAB so it can easily be run for alternate cases. Part (d) Plot the horizontal component of the force exerted on each cylinder for different ranges of acceleration, which illustrate the various case of rolling and sliding obtained in parts (a) and (b).

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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