c
Frictional force,
${F=\mu R=0.5 \times m g=0.5 \times 60=30 \mathrm{N}}$
$\mathrm{Now}$ $\mathrm{F}=\mathrm{T}_{1}=\mathrm{T}_{2} \cos 45^{\circ}$
$\text { or } \quad 30=\mathrm{T}_{2} \cos 45^{\circ}$
${\text { and }} {W=T_{2} \sin 45^{\circ}} $
${\therefore \quad} {W=30 \mathrm{N}}$