GATE ME 2020 Set 2 — Question 30

Let $P_x$ and $P_y$ be the horizontal and vertical reaction forces at the hinge P, respectively. Let $Q_x$ be the horizontal reaction force at the roller support Q (since the roller is on a vertical surface, the reaction is horizontal).The system is in static equilibrium. To find the reaction force at Q, we can take the sum of moments about the hinge P and set it to zero.$\sum M_P = 0$Let's consider counter-clockwise moments as positive and clockwise moments as negative.The applied load of 1200 N at point R creates a clockwise moment about P. The perpendicular distance from P to the line of action of the load is 5 m.
Moment due to load = $-1200 \text{ N} \times 5 \text{ m} = -6000 \text{ Nm}$.The reaction force $Q_x$ at support Q creates a counter-clockwise moment about P. The perpendicular distance from P to the line of action of $Q_x$ is 4 m.
Moment due to reaction at Q = $+Q_x \times 4 \text{ m}$.The reaction forces at P ($P_x$ and $P_y$) pass through point P, so they do not create any moment about P.Summing the moments:
$Q_x \times 4 - 1200 \times 5 = 0$
$4 Q_x = 6000$
$Q_x = \frac{6000}{4} = 1500 \text{ N}$Thus, the magnitude of the reaction force at support Q is 1500 N.