Question

Neglecting the weight of the beam below find the tension in tie rope and the force components at the hinge in the terms of w.repeat if the uniform beam weighs 1/2W

Answer 1

First case:-

Let weight of the beam be negligible.

The beam is in equilibrium so we can equate net torque acting on it about the hinge to zero.

Torque due to the weight W about hinge $\sf{=-0.8WL\sin50^o}$

Negative sign is due to clockwise rotational effect.

Torque due to tension in the string about hinge $\sf{=TL\sin20^o}$

Then, net torque is zero.

$\longrightarrow\sf{TL\sin20^o-0.8WL\sin50^o=0}$

$\longrightarrow\sf{T=\dfrac{0.8W\sin50^o}{\sin20^o}}$

$\longrightarrow\sf{T=\dfrac{0.8W\times0.766}{0.342}}$

$\longrightarrow\sf{\underline{\underline{T=1.792W}}}$

Equate the net horizontal force to zero since the beam is in equilibrium.

$\longrightarrow\sf{F_H-T\sin70^o=0}$

$\longrightarrow\sf{F_H=T\sin70^o}$

$\longrightarrow\sf{F_H=1.792W\times0.940}$

$\longrightarrow\sf{\underline{\underline{F_H=1.684W}}}$

Equate net vertical force to zero similarly.

$\longrightarrow\sf{F_V-W-T\cos70^o=0}$

$\longrightarrow\sf{F_V=W+T\cos70^o}$

$\longrightarrow\sf{F_V=W+1.792W\times0.342}$

$\longrightarrow\sf{\underline{\underline{F_V=1.613W}}}$

Second case:-

Let the weight of the beam be $\sf{\dfrac{W}{2}.}$

The torque due to weight of the beam about hinge $\sf{=-0.5L\cdot\dfrac{W}{2}\sin50^o=-0.25WL\sin50^o}$

So the net torque is zero.

$\longrightarrow\sf{TL\sin20^o-0.8WL\sin50^o-0.25WL\sin50^o=0}$

$\longrightarrow\sf{T\sin20^o=1.05W\sin50^o}$

$\longrightarrow\sf{T=\dfrac{1.05W\sin50^o}{\sin20^o}}$

$\longrightarrow\sf{T=\dfrac{1.05W\times0.766}{0.342}}$

$\longrightarrow\sf{\underline{\underline{T=2.352W}}}$

The net horizontal force doesn't change and remains the same as that in first case, since weight of the beam is acting vertically downwards. Therefore,

$\longrightarrow\sf{\underline{\underline{F_H=1.684W}}}$

Equate net vertical force to zero.

$\longrightarrow\sf{F_V-\dfrac{W}{2}-W-T\cos70^o=0}$

$\longrightarrow\sf{F_V=\dfrac{3W}{2}+T\cos70^o}$

$\longrightarrow\sf{F_V=1.5W+2.352W\times0.342}$

$\longrightarrow\sf{\underline{\underline{F_V=2.304W}}}$