Question

In the system as shown in figure $AB$ is a uniform rod of mass $10kg$ and $BC$ is a light string which is connected between wall and rod, in vertical plane. There is block of mass $15kg$ connected at $B$ with a light string.
(Take $g=10m/s^2$) ($BC$ and $BD$ are two different strings)
If whole of the system is in equilibrium then find
(i) Tension in the string $BC$
(ii) Hinge force exerted on beam at point $A$

Answer 1

Since the system is in equilibrium;

The moment of forces about point A should be zero.

Thus;

$10g\times 0.5+15g\times 1=Tsin{53}^o\times 150+150=0.8TT=250N$

At point A, the hinge force is working which can be resolved as $R_y\&R_x$

$R_x=Tcos{53}^o{R}_x=250\times 0.6=150N{R}_y=Tsin{53}^o{R}_y=250\times 0.8=200N$

$R=\sqrt{(R_x{)}^2+(R_y{)}^2}R=\sqrt{{150}^2+{200}^2}R=250N$

Therefore, The tension in the string will be $250N$ and the hinged force at point A will be $250N$