Answer
(a) $0.255KN$
(b) $0.51KN$
(c) $0.51KN$
Work Step by Step
(a) We can find the required force as
$\Sigma F_y=F+F-C_{lower}=0$
$\implies 2F=C_{lower}$
$\implies F=\frac{1}{2}C_{lower}$
$\implies F=\frac{1}{2}mg$
We plug in the known values to obtain:
$F=\frac{1}{2}(52)(9.81)$
$F=255N=0.255KN$
(b) We can find the tension in the upper chain as
$\Sigma F_y=C_{upper}-(F+F)=0$
$\implies C_{upper}=2F$
We plug in the known values to obtain:
$C_{upper}=2(0.255KN)$
$C_{upper}=0.51KN$
(c) We can find the tension in the lower chain as follows:
$C_{lower}=mg$
We plug in the known values to obtain:
$C_{lower}=(52)(9.81)$
$C_{lower}=510N=0.51KN$
