Question

A square block of wood of mass $m$ is hinged at A and rests on a roller at B. It is pulled by means of a string attached at D and inclined at an angle of ${30}^0$ with the horizontal, as shown in the figure. The force P which should be applied to the string to just lift the block off the roller is approximately $\left(\frac{Mg}{n}\right)$. Find the value of $n$. (Given $sin75$${}^0$$=0.9660$)

• A. $3$
• B. $2$
• C. $4$
• D. $1$

Answer 1

Answer: (A)

$3$

The force 'P' exerted on block by the string is balanced due to its weight 'Mg', and the reactions 'Ra' and 'Rb' at A and B respectively.
Now, when the block is just off the roller at B, $Rb=0$
since, Ra is unknown in deflection it can be eliminated by taking moments about A.
for equilibrium,
$\sum Ra=0$
Now joining the lines AD and AE as shown in figure, we get the equation for forces as,
$Mg\left(\frac{a}{2}\right)=P\left(AD\sin {75}^{\circ }\right)$
from figure,
$AD=\sqrt{(A{E}^2+E{D}^2)}$
$AD=a\sqrt{2}$
using this value in above equation, we get
$Mg\left(\frac{a}{2}\right)=P\left(a\sqrt{2}\right)\sin {75}^{\circ }$
$Mg\left(\frac{a}{2}\right)=P\left(a\sqrt{2}\right)\left(0.966\right)$
$P=\left(0.37\right)Mg$
$P=\left(\frac{1}{3}\right)Mg$
But the approximate force that should be applied to the string to lift the block of the roller is $\left(\frac{1}{n}\right)Mg$.
therefore,
$n=3$