The Requirements of Equilibrium — Question 6

Question 6

An applied couple redistributes support reactions without changing total weight

A weightless horizontal beam spans $4\,\mathrm m$ between an ideal pin at $A$ and a vertical roller at $B$. A $100\,\mathrm N$ load acts downward $1\,\mathrm m$ from $A$, and a $20\,\mathrm{N\,m}$ counterclockwise couple is applied.

1. Find all support reaction components.

2. Check the moment equation about $B$ independently.

Question 6 - Solution

An applied couple redistributes support reactions without changing total weight

(a) Derivation and calculation

Take upward and counterclockwise positive. Balance gives $$\sum F_x=A_x=0,\qquad A_y+B_y=100.$$ About $A$, $$4B_y-100(1)+20=0\quad\Longrightarrow\quad\boxed{B_y=20\ \mathrm N}.$$ Thus $$\boxed{A_y=80\ \mathrm N},\qquad\boxed{A_x=0}.$$

(b) Result and interpretation

Taking moments independently about $B$, $$\sum\tau_B=-4A_y+100(3)+20 =-4(80)+300+20=0.$$ Without the couple, $B_y=25$ N and $A_y=75$ N. The applied $20$ N m couple changes them by $$\Delta B_y=-20/4=-5\ \mathrm N,\quad\Delta A_y=+5\ \mathrm N.$$

Physical check

The total upward reaction remains 100 N because a pure couple adds no net force.