Question

Question 2. The all-wheel drive car shown in Figure 2 is at rest on a level road. The car's mass centre is at G. The car starts moving with an initial acceleration. For this car: (a) Draw the free body diagram showing all the forces acting on the car. (b) Assuming the coefficient of static friction between the tyres and the track is, for front tyres and Hz for rear tyres, obtain an expression for the maximum possible initial acceleration that can be achieved by the car (Qmax) without having any of the wheels slip on the track. (C) Show that if f: = 72 = 4, the expression obtained in (b) simplifies to amax [3 marks] [15 marks] [2 marks] Neglect the mass of the wheels [20 marks] (1) Design part (using MATLAB): You have two pairs of tyres, type A with the coefficient of static friction of 1.1 and type B with the coefficient of static friction of 0.8. Assuming h = 0.5 m and 12 + 12 = 2.6 m, discuss in detail which type should be used for front wheels and which one for the rear wheels to achieve the highest maximum initial acceleration? Develop a complete design, considering both possible cases (i.e. using type A for front and B for rear and vice versa). Consider las a parameter and vary it in the range [0 2.6) and plot amar versus 1/4 + 1x), and compare it to the case of part (e). Explain each part clearly and come up with a final optimum design with realistic values for l, and lz. Total: [40 marks R 12 11 Figure 2 Page 4 of 5

Answer 1

* All wheel drive. Direction of Motion ma h ng 1 H₂ 2 fz e Hi R F l,th * Reautions: (R. F:). At Dynamic equilibicam, SMR=0 FX(1, +22) + maxh = mg x la Fa mg l -mah litla IMF 20 RX(d,+ 12) = mantinge, R= mge, tman litda * limiting friction : fiaMif - (mgl-mah) Mi litla +2 - H2R = (mge,+ man) Hz litla

* If the applied frictional force is maximum limiting Fractional these as compare with occure Force then slip would At dynamic equilibrium. EFx=0 fit fa= ma h) Hi ma + litle (ongda-mah inge, tmah) H₂ litla (gez-ahlf, +(getah)/2 litla Hal-kah tHage, tha ah = all,+*2) = a. Higle+H2gl, = all,+2 +4 h-42b) Higdz + Hage, the ar litlaten-Hah. Here The acceleration fle, &2, h, Hirta) g-only constant as

c) Hi=42=4 regl, trige, a litlatfh-feh ugllitle) aa Elreg. d) litla caseli Front ₃ type A tyre HisHA = 11 Rear -> type B type H2=HB=0.8: from matlab result: is zero when the front wheel reaction li 7.848 mis? e = 0.8462 l, 0.8462x 2.6 di = 212, lao.em At Ta=1012.mis, the position of co dico l2=2.6m. possible acceleration This is the maxincem in case :

case 2: Front → type B tyre His MB=0.8 Rear → type A tyre H₂=HA=k from matlab result: when the front wheel reuetoon 2eno ł -0.7885 lis0.7881* 2.6 a = 10.19 m/s² lis 2.05m lz=0.55 m. ه ما the position / At, a= 8.328 m/s² of li T 2,5-0 22=2.6m conchesion * Front wheel tyre should have type * Rear wheel tyre should have a type. Max. acceleration : 8-10.19m/s? 2.05%, 22=0.55m, h 20.5m. *l- type A type B. Osm lz Kassm li 2.0 sm

Case1:Front-->A type Rear-->B type 10.5 Case2:Front--> type Rear-->A type 11.5 10 7 10 11 8 6 9.5 5 10.5 6 9 10 acceleration in m/sec2 Front wheel reaction acceleration in m/sec2 Front wheel reaction 8.5 9.5 8 9 10 7.5 -1 8.5 -2 -2 0 0.1 0.2 0.3 0.4 0.7 0.8 0.9 1 8 -4 0.5 0.6 11/(11+12) 0 0.1 0.2 0.3 0.7 0.8 0.9 1 0.4 0.5 0.6 11/(11+12) Case1:Front-->A type Rear--> type 10.5 X: 0 Y: 10.2 Case2:Front--> type Rear-->A type 8 11.5 10 7 10 11 X: 0.7885 Y: 10.79 18 6 9.5 10.5 16 9 10 acceleration in m/sec2 Front wheel reaction acceleration in m/sec Front wheel reaction 8.5 9.5 X: 0.8462 Y: 7.848 1 X: 0.7885 Y: 1.025e-15 8 X: 0.8462 Y: 0 9 0 7.5 8.5 -2 7 8 X: 0 Y: 8.328 0 -2 1 0.1 0.2 0.3 0.7 0.8 0.9 0 0.2 -4 1 0.3 0.4 0.7 0.4 0.5 0.6 11/(11+12) 0.8 0.9 0.5 0.6 11/(11+12)

clc;clear all;close all;
h=0.5;
l=2.6;
muA=1.1;
muB=0.8;
g=9.81;
l1=0:0.01:2.6;
l2=l-l1;
%case1 type A tyre is front and type B tyre is rear
mu1=muA;
mu2=muB;
a=((mu1*g*l2)+(mu2*g*l1))./(l1+l2+(mu1*h)-(mu2*h));
F=(g*l2-a*h)./(l1+l2);
figure
yyaxis left
plot(l1./l,a)
xlabel('l1/(l1+l2)')
ylabel('acceleration in m/sec^2')
yyaxis right
plot(l1./l,F)
ylabel('Front wheel reaction')
title('Case1:Front-->A type Rear-->B type')

%case2 type B tyre is front and type A tyre is rear
mu1=muB;
mu2=muA;
a=((mu1*g*l2)+(mu2*g*l1))./(l1+l2+(mu1*h)-(mu2*h));
F=(g*l2-a*h)./(l1+l2);
figure
yyaxis left
plot(l1./l,a)
xlabel('l1/(l1+l2)')
ylabel('acceleration in m/sec^2')
yyaxis right
plot(l1./l,F)
ylabel('Front wheel reaction')
title('Case2:Front-->B type Rear-->A type')