A $\mathbf{73}\mathbf{\text{\hspace{0.17em}kg}}$ man stands on a level bridge of length $\mathit{L}$. He is at distance$\mathit{L}\mathbf{/}\mathbf{4}$ from one end. The bridge is uniform and weighs$\mathbf{2}\mathbf{.7}\mathbf{\text{\hspace{0.17em}kN}}$ . What are the magnitudes of the vertical forces on the bridge from its supports at (a) the end farther from him and (b) the nearer end?

The lengthof the bridgeis L.

The weight of the beam is$2600\text{\hspace{0.17em}N}$

The weight of a man is $715.5\text{\hspace{0.17em}N}$

Using the condition for the static equilibrium, you can write the equation for the torque and the equation for the force. Using these equations, you can solve the unknown forces.

The equations are given below.

$\begin{array}{l}\sum F_x=0\\ \sum F_y=0\\ \sum \tau =0\end{array}$

From the given diagram,

Let F_a and F_b are vertical forces from two different ends.

Man is at distance$\text{L/4}$having weight$715.4\text{\hspace{0.17em}N}$

As the bridge is uniform, its weight at the middle point$\text{L/2}$and weight$2700\text{\hspace{0.17em}N}$

From the equilibrium condition, we can write

$\begin{array}{l}{F}_a+F_b-2700\text{\hspace{0.17em}N}-715.4\text{\hspace{0.17em}N}=0\\ F_a+F_b=3415.4\text{\hspace{0.17em}N}\end{array}$

We also can write,

$\begin{array}{l}(F_b\times L)-(2700\times \frac{L}{2})-(715.4\times \frac{L}{4})=0\\ F_b-1350N-178.8N=0\\ F_b=1530\text{\hspace{0.17em}N}\end{array}$

Therefore, we can get from the above expressions that, $F_a=1885.4\text{\hspace{0.17em}N}$$F_a=1885.4\text{\hspace{0.17em}N}$$F_a=1885.4\text{\hspace{0.17em}N}$