Question

p
l
e
a
s
e
answer
i
t
i
s
a
j
e
e
q
u
e
s
t
i
o
n
2
0
1
8​

Answer 1

Answer:-

$\displaystyle\Large\boxed {\sf {(c)\ \tan\theta=\dfrac {1}{2}}}$

Solution:-

We know that moment of force,

$\displaystyle\longrightarrow\sf{\tau=Fr\sin\alpha}$

where r is the separation of the applied force F about point A and $\displaystyle\sf {\alpha}$ is the angle between F and r.

Here,

$\displaystyle\longrightarrow\sf{r=AB}$

Since the angle of F with the horizontal is $\displaystyle\sf {\theta}$ and the angle between F and r is $\displaystyle\sf {\alpha,}$ then the angle (acute) of r = AB with the horizontal will be $\displaystyle\sf {(180^o-\alpha-\theta)}$ and is given by,

$\displaystyle\longrightarrow\sf{\tan (180^o-\alpha-\theta)=\dfrac {4}{2}}$

$\displaystyle\longrightarrow\sf{\tan (180^o-(\alpha+\theta))=2}$

Since $\displaystyle\sf {\tan (180^o-x)=-\tan x,}$

$\displaystyle\longrightarrow\sf{-\tan (\alpha+\theta)=2\quad\quad\dots (1)}$

Now, consider the moment of force,

$\displaystyle\longrightarrow\sf{\tau=Fr\sin\alpha}$

For maximum value of $\displaystyle\sf {\tau,\ \sin\alpha}$ should be maximum. Thus,

$\displaystyle\longrightarrow\sf{\sin\alpha=1}$

Since $\displaystyle\sf {0^o\leq\alpha\leq90^o,}$

$\displaystyle\longrightarrow\sf{\alpha=90^o}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{-\tan (90^o+\theta)=2}$

Since $\displaystyle\sf {\tan (90^o+x)=-\cot x,}$

$\displaystyle\longrightarrow\sf{\cot\theta=2}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\tan\theta=\dfrac {1}{2}}}}$

Hence (c) is the answer.