Question

A pin jointed uniform rigid rod of weight W and length L is supported horizontally by an external force F as shown in the figure below. The force F is suddenly removed. At the instant of force removal, the magnitude of vertical reaction developed at the support is

• A. Zero
• B. $\frac{W}{4}$
• C. $\frac{W}{2}$
• D. W

Answer 1

The correct option is B

$\frac{W}{4}$

$\sum R_x=0\Rightarrow R_x=0$

$\sum R_y=0\Rightarrow R_y+F=W$

$R_y=(W-F)$

If force F removes, then rod will rotate then

$T_{\left(\text{torque}\right)}=I\alpha$

$W\times \frac{L}{2}=\left(\frac{W}{g}\right)\frac{L^3}{3}\times \alpha ....\left(ii\right)$

$W-R_y=\left(\frac{W}{g}\right)a_{cg}....\left(iii\right)$

$a_{cg}=\left(\frac{L}{2}\right)\alpha .....\left(iv\right)$

Equations (ii), (iii), (iv)

$W\times \frac{L}{2}=\left(\frac{W}{g}\right)\frac{L^2}{3}\times \frac{a_{c.m}}{\left(\frac{L}{2}\right)}$

$W\times \frac{L}{2}=\left(\frac{W}{g}\right)\times \frac{L^2}{3}\times \left(\frac{2}{L}\right)\times \frac{(W-R_y)}{\left(\frac{W}{g}\right)}$

$\frac{3W}{4}=W-R_y$

$R_y=\frac{W}{4}$