Answer
(a) $1.96KN$
(b) $2.6KN$
Work Step by Step
We know that the net torque is zero
$\implies Tcos25^{\circ}\times 4.55-N\times 11.6=0$
This simplifies to:
$T=\frac{N\times 11.6}{4.55cos25^{\circ}}=\frac{697\times 11.6}{4.55cos25^{\circ}}=1.96KN$
(b) In the given case $\sum F_x=0$
$\implies Tsin25^{\circ}=F_{cx}$
$\implies F_{cx}=1.96\times sin25=829N$
and $\sum F_y=0$
$\implies Tcos25^{\circ}+N=F_{cy}$
$F_{cy}=1.96\times cos25^{\circ}+697$
$F_{cy}=1.96cos25^{\circ}+697=2.5KN$
Now $F_c=\sqrt{F_{cx}^2+F_{cy}^2}$
We plug in the known values to obtain:
$F_c=\sqrt{(829)^2+(2.5)^2}=2.6KN$
