Solve the following :

(a)

Velocity of cyclist $=\frac{(18 \times 5)}{18}=5 \mathrm{~m} / \mathrm{s}$

At point B

$\mathrm{N}_{\mathrm{B}}+\frac{m v^{2}}{R}=\mathrm{mg}$

$N_{B}=(100)(10)-\frac{(100)(5)^{2}}{100}$

$N_{B}=(100)(10)-$

$\mathrm{N}_{\mathrm{B}}=975 \mathrm{~N}$

At point $D$

$\mathrm{N}_{\mathrm{D}}=\mathrm{mg}+\frac{m v^{2}}{R}$

$N_{D}=(100)(10)+\frac{(100)(5)^{2}}{100}$

$N_{D}=1025 \mathrm{~N}$

(b)

No force acts along surface at point B and D

So, $\mathrm{ff}_{\mathrm{B}}=\mathrm{ff}_{\mathrm{D}}=0$

At point $C$,

$\because$ acceleration $=0$

$\mathrm{So}, \mathrm{ffc}=\mathrm{mg} \sin \theta$

$f f c=(100)(10) \sin 450$

$f f c=707 N$

(C)

Before C,

$\mathrm{N}_{\mathrm{c}}+\frac{m V^{2}}{R}=\operatorname{mgcos} 45^{\circ}$

$\mathrm{N}_{\mathrm{c}}=(100)(10)\left(\frac{1}{\sqrt{2}}\right)-\frac{(100)(5)^{2}}{(100)}$

$\mathrm{N}_{\mathrm{C}}=682 \mathrm{~N}$

After C,

$\mathrm{N}_{\mathrm{C}}=\frac{m V^{2}}{R}+\mathrm{mg} \cos 45^{\circ}$

$\mathrm{N}_{\mathrm{C}}=\frac{(100)(5)^{2}}{100}+(100)(10)\left(\frac{1}{\sqrt{2}}\right)$

$\mathrm{N}_{\mathrm{C}}^{\prime}=732 \mathrm{~N}$

(d) $f f=\mu N$

For minimum friction force normal contact should be minimum.

Normal contact is minimum at point just before C.

$\mathrm{ff}=\mathrm{mg} \sin \theta=\mu \mathrm{N}$

$(100)(10) \sin 45^{\circ}=\mu(682)$

$\mu=1.037$