Question

Forces of magnitudes $3,P,5,10$ and $Q$ are respectively acting along the sides $AB,BC,CD,AD$ and the diagonal $CA$ of a rectangle $ABCD$, where $AB=4m$ and $BC=3m$. If the resultant is a single force along the other diagonal $BD$, then $P,Q$ and the resultant are

• A. $4,10\frac{5}{12},12\frac{11}{12}$
• B. $5,6,7$
• C. $3\frac{1}{2}$,$8$, $9\frac{1}{2}$
• D. None of the above

Answer 1

The correct option is A $4,10\frac{5}{12},12\frac{11}{12}$

$ABCD$ is a rectangle in which $AB=4m$ and $BC=3m$
Then, $\tan \theta =\frac{BC}{AB}=\frac{3}{4}$
The forces $3,P,5,10$ and $Q$ newtons have the resultant $R$ Newton as shown in the figure.
$R\cos \theta +5-3$
$=Q\cos \theta +2....\left(i\right)$
and $R\sin \theta =P+10-Q\sin \theta ....\left(ii\right)$
and $QAB\sin \theta +5.BC=10.AB$
or $Q4.\left(3/5\right)+5\times 3=10\times 4(\because \sin \theta =\frac{3}{5}and\cos \theta =\frac{4}{5})$
$\Rightarrow \frac{12}{5}Q=40-15=25$
$\Rightarrow Q=\frac{125}{12}newton$
Then, find $R$ from Eq. (i) and $P$ from Eq. (ii).
$\frac{4R}{5}=\frac{125}{12}\times \frac{4}{5}+2$
$\Rightarrow 4R=\frac{125}{3}+10=\frac{155}{3}$
$\Rightarrow R=\frac{155}{12}$
and $\frac{155}{12}\times \frac{3}{5}=P+10-\frac{125}{12}\times \frac{3}{5}$
$\frac{155}{4}=5P+50-\frac{125}{4}$
$\Rightarrow 5P=\frac{155}{4}+\frac{125}{4}-50$
$\Rightarrow 5P=\frac{280}{4}-50=70-50=20$
$\Rightarrow P=4$.