Question

Consider the arrangement shown in the figure. Assuming friction-less contacts, then the magnitude of external horizontal force $P$ applied at the lower end for an equilibrium of the rod will be: (The rod is uniform and its mass is ${}^{\prime }{m}^{\prime }$)

• A. $\frac{mg}{2}$
• B. $\frac{mg}{2}\cot \theta$
• C. $\frac{mg}{2}\tan \theta$
• D. $\frac{mg}{2}\sec \theta$

Answer 1

The correct option is B $\frac{mg}{2}\cot \theta$

Let length of the
$rod=l$
from $\sum f_x=0$
$P=R_2$
from $\sum f_y=0$
$R_1=mg$
taking moment at $A$
$R_2\times BC=mg\times AM$
$R_2\times L\sin \theta =mg\times \frac{1}{2}\cos \theta$
$R_2=\frac{mg}{2}\frac{l\cos \theta }{\sin \theta }$
$R_2=P=\frac{mg}{2}\cot \theta$