Answer to Question #135882 in Mechanics | Relativity for Amita

Question

A uniform rod of mass 200g and length,L is supported at both ends.Masses of 300g and 500g are suspended at L/3 and 3L/4 from one end. Determine the force exerted on each support.

Answer 1

From the principle of moments,

sum of upward forces = sum of downward forces

moment $=$ force($F$) $\times$ distance($L$)

moment $=$ $F \times L$

Taking moment about one end A we have,

moment $=$ $\frac{500}{100} \times \frac{3L}{4} + \frac{200}{100} \times \frac{1L}{2} + \frac{300}{100} \times \frac{1L}{3}$

$=$ $3.75L + 1L + 1L = 5.75L$

The multiplier, $10^{-2}$ or division by $100$, is done to convert the given masses from grams to newtons.

moment about A $= 5.75L = F_{1} \times L$

$F_{1} = 5.75N$

sum of downward forces $= \frac{500N}{100} + \frac{200N}{100} + \frac{300N}{100}$

$5N + 2N + 3N = 10N$

Therefore, force, $F_{2}$ on the other end B will be,

$F_{2} = 10N - 5.75N = 4.25N$