Question

A long uniform beam is acted upon by several forces as shown in figure, and it is in complete equilibrium. What will be the value of forces $X\&Y$? Neglect the mass of the beam.

• A. $X=3\text{N},Y=3\text{N}$
• B. $X=0,Y=6\text{N}$
• C. $X=6\text{N},Y=3\text{N}$
• D. $X=6\text{N},Y=6\text{N}$

Answer 1

The correct option is D $X=6\text{N},Y=6\text{N}$


For the complete equilibrium of the body, net external force as well as net external torque should be zero.

For translational equilibrium of body,
$F_{\text{net}}=0$
Considering upwards direction as $+ve$
$\sum F=0$
$\Rightarrow Y+6-6-X=0$
$\therefore Y=X...\left(i\right)$
For rotational equilibrium, the net torque be zero about point $B$:
${\tau }_{\text{net}}=0...\left(ii\right)$
$\tau =F\times r_{\perp }$
$\because$Force $Y$ passes through reference point $B$, hence torque ${\tau }_Y=0$ about point $B$.
Taking anticlockwise sense of rotation for torque as $+ve$ and substituting the torque with proper sign in Eq $\left(ii\right)$,
$\sum \tau =0$
$\Rightarrow +X\left(30\right)+0+6\left(10\right)-6\left(40\right)=0$
$\therefore X=\frac{180}{30}=6\text{N}$
From Eq $\left(i\right)$,$X=Y$
$\Rightarrow Y=6\text{N}$