91Ó°ÊÓ

The length of a beam is L.

The load of the beam is equally divided by three men.

Using the equation for the equilibrium of force and the equation of equilibrium of torque, you can solve for the distance of a crosspiece. The equations are given below.

At equilibrium,${\mathbf{F}}_{\mathbf{net}}\mathbf{=}\mathbf{0}$

At equilibrium,${\mathbf{Ï„}}_{\mathbf{net}}\mathbf{=}\mathbf{0}$

You have given that the load of the beam is equally divided by three men.

Suppose, one man is applying upward force F.

At equilibrium,

${\text{F}}_{\text{net}}=0$

So,

role="math" localid="1661360361518" $\text{W}=3×\text{¹ó â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\left(1\right)$

Where W is the weight of the beam.

Now, let us consider the pivot point, at the man of one end. Then the equation of equilibrium of torque can be written as,

At equilibrium,

${\text{Ï„}}_{\text{net}}=0$

So,

$\text{W}×\frac{\text{L}}{\text{2}}−2\text{F}×\text{x}=0$

Solving this equation for the distance of the crosspiece from the man at one end, you will get,

$\mathrm{x}=\frac{\mathrm{W}×\mathrm{L}}{2×2\mathrm{F}}$

Now, substituting eq. (1)

$\mathrm{x}=\frac{3}{4}\mathrm{L}$

So, the distance of the crosspiece from the free end is,

$\begin{array}{c}\mathrm{d}=\mathrm{L}−\mathrm{x}\\ \mathrm{d}=\frac{\mathrm{L}}{4}\end{array}$

Thus, the distance of the crosspiece from the free end of the beam is, $\text{d}=\frac{\text{L}}{\text{4}}$.