Question

A person carries a hammer on his shoulder and holds the other end of its light handle in his hand. Let $y$ be the distance between his hand and the point of support. If the person changes $y$, the pressure on his hand will be proportional to:

• A. $y$
• B. $y^2$
• C. $\frac{1}{y}$
• D. $\frac{1}{y^2}$

Answer 1

Answer: (C)

$\frac{1}{y}$

As it is clear from the Free body diagram, for the hammer to be in equilibrium condition following should be followed :
$F\times y=w\times (l-y)$
which gives;
$F\times y=w\times l-w\times y$
$y=\frac{w\times l}{F+w}$
$\frac{F+w}{w\times l}=\frac{1}{y}$
Let $w\times l=d$ and $\frac{1}{l}=c$
Hence the new equation becomes;
$\frac{F}{d}+c=\frac{1}{y}$
which gives $F=\frac{d}{y}-c\times d$
Let $c\times d=a$ , which gives
$F+a=\frac{d}{y}$
As "$a$" being a constant does not effect proportionality we have $F$ inversely proportional to $y$.