M= 1800 kg
X= 1.6 m
R= 24 m
V$= 36 kmph \\
= 36× \dfrac{5}{18}$
∴V = 10 m/sec.
dw= 600 mm = 0.6 m
rw= 0.3 m
h= 0.95 m
Mass of the wheel= 180 kg
Radius of gyration $= 240 mm \\
= 0.24 m$
Considering
First conndering the effect of (w=mg) of the car and of the centrifugal force, on it determining the Reactions RA and RB Ref. above fig. 1
Resolving the 1hr forces to the trag.
RA+RB $= Mg \cos ѳ +\dfrac{ Mv^2}{R} \sin θ \\
= 1800×9.81×\cos 10°+\dfrac{1800(10)^2}{24}×\sin 10°$
$\boxed{RA+RB= 18692 N}$
Now taking moment @ B.
$RA×W= Mg \cos θ×\dfrac{W}{2} + Fe \sin θ× \dfrac{W}{2} + Mg \sin θ×h ─ Fc \cos θ×h$
$RA= [Mg \cos ѳ+\dfrac{Mr^2}{R} \sin ѳ] × ½ + [Mg \sin ѳ– \dfrac{Mr^2}{R} \cos ѳ]×\dfrac{h}{w}$
$RA = [1800×9.81×\cos 10+1800×\dfrac{10^2}{24}× \sin 10]×\dfrac{1}{2}
+[1800×9.81×\sin 10 ─(1800×\dfrac{10^2}{24}× \cos 10]×\dfrac{0.95}{1.6}$
RA= 6781N
RB= 11911N
Reactions due to gyroscopic couples.
Cw$= 2 Iw×Wω× \cosθ×wp \\
= 2×mk^2×\dfrac{V^2}{rR} × \cosθ \\
= 2×180×(0.24)^2× \dfrac{10^2}{(0.3×24)}× \cos10 \\
= 283.6 N─M$
Reactions on outer wheel,
$R_{G \ \ outer}= \dfrac{C_W}{2_W} = \dfrac{283.6}{(2×1.6 )}= 88.6 N$ upwards(↑)
$R_{G \ \ inner} = 88.6 N downwards(↓)$
Pressure on each outer rails= RB + RG outer
= 11911+88.6=11999.6 (downwards)
Similarly on each inner wheel $= R_A─ R_{G \ \ inner} \\
= 6781─ 88.6= 6692.4 N (upwards)$