Explanation:
The first step is to write the 1st condition of equilibrium, which states that the net force acting on the beam is zero:
[tex]F_{net} = 5\:\text{kN} + N_L + N_R - 40\:\text{kN} - 60\:\text{kN} = 0[/tex]
where [tex]N_L[/tex] and [tex]N_R[/tex] are reaction forces at the beam supports on the left and right, respectively.
Next, the 2nd condition of equilibrium states that the net torque about a pivot point is zero. Let's choose the location of the left beam support as our pivot point, which will make the torque due to [tex]N_L[/tex] is equal to zero. Let us also assume that the counterclockwise direction produces positive torque. So we can write
[tex]\tau_{net} = (40\:\text{kN})(2\:\text{m}) + (5\:\text{kN})(5\:\text{m}) \\+ N_R(7\:\text{m}) - (60\:\text{kN})(8\:\text{m}) = 0[/tex]
[tex]\Rightarrow 80\:\text{kN-m} + 25\:\text{kN-m} + N_R(7\:\text{m}) \\= 480\:\text{kN-m}[/tex]
Solving for [tex]N_R,[/tex] we see that
[tex]N_R = 53.6\:\text{kN}[/tex]
Putting this value into our 1st equation, we find that the reaction force [tex]N_L[/tex] is
[tex]N_L = 100\:\text{kN} - 5\:\text{kN} - N_R[/tex]
[tex]\:\:\:\:\:\:\:\:= 41.4\:\text{kN}[/tex]
