Question

A uniform rod is $4m$ long and weight $10Kg$. If it is supported on a knife edge at one metre from the end, what weight placed at that end keeps the rod horizontal?

• A. $8N$
• B. $10N$
• C. $12N$
• D. $15N$

Answer 1

Answer: (B)

$10N$

Linear density of rod, $\mu =\frac{M}{L}=\frac{10}{4}=2.5kg{m}^{-1}$

Weight, $W$

$W_{AB}=\mu \times 1=2.5kg$

$W_{BC}=\mu \times 3=2.5\times 3=7.5kg$

Center of mass of AB from knife is $0.5m$

Center of mass of BC from knife is $1.5m$

Length of AB is short, so force apply at point A to balance the rod

Moment at point B

$F\times L_{AB}=W_{BC}\times 1.5-W_{AB}\times 0.5$

$F=\frac{W_{BC}\times 1.5-W_{AB}\times 0.5}{L_{AB}}=\frac{7.5\times 1.5+2.5\times 0.5}{1}$

$F=10N$

Weight placed at end of rod is $10N$