Question

In figure the uniform gate weights 300 N and is 3 m wide and 2 m high. It is supported by a hinge at the bottom left corner and a horizontal cable at the top left corner, as shown. Find:
(a) The tension in the cable
(b) The force that the hinge exerts on the gate ( magnitude and direction ).

Answer 1

C is the centre of mass of gate.

So position C is $(\frac{9}{2},\frac{2}{2})=(1.5,1)$

The tension of string is T & hinge exert a force F on gate.

Components of F are $F_h\&F_v$ as shown in figure.

The component $F_h$ and tension of string are equal as it will produce couple about axis AB.

$T=F_h--------\left(1\right)$

And vertical component of force F will balance the weight of gate that is

$F_v=300N--------\left(2\right)$

Now as the gate doesnot rotate about point B , the torque about point B is zero.

$-T\times 3+F_v\times 0+F_h\times 0+300\times 1.5=0$

$⟹T=150N$

$\left(b\right)$ Hence $F_h=150N,F_v=300N$

$F=\sqrt{F_h^2+F_v^2}=\sqrt{{150}^2+{300}^2}$

$⟹335N$ (Direction is given in figure)