Answer to Question #87002 in Molecular Physics | Thermodynamics for Ezekiel Fred

Question

A uniform metre rule of weight 1.0N is pivoted at the 0.40m mark. A 2.0N weight is hung from the 0.15m mark. Where must a 2.0N weight be placed to balance the rule?

Answer 1

$$M_1 =M_2$$

$M_1 =P_1 \times (x_0-x_1) + P_l(x_0 - x_l)$

$P_1 = 2 N$

$x_o = 0.4 m$

$x_l = \frac{x_0}{2} = \frac{0.4 m}{2} = 0.2 m$

$P_l = P\times \frac{x_0}{L} = 1 N \times \frac {0.4 m}{1 m} = 0.4 N$

$M_1 = 2N \times (0.4 m - 0.15 m) + 0.4 N \times (0.4 m - 0.2 m) = 0.58 N \times m$

$M_2 = P_2 \times ( x_2 - x_0) + P_r ( x_r - x_0)$

$P_2 = 2N$

$P_r = P \times \frac{L-x_0}{L} = 1 N \times \frac{1 m -0.4 m}{1 m} = 0.6 N$

$x_r = x_0 + \frac{L-x_0}{2} = 0.4 m + \frac{1 m - 0.4 m}{2} = 0.7 m$

$M_2 = 2 N \times (x_2 -0.4 m) + 0.6 N \times (0.7 m - 0.4 m ) = 2N \times (x_2 - 0.4 m) + 0.18 N\times m$

$M_1 =M_2$

$0.58 N\times m = 2 N \times (x_2 -0.4 m) + 0.18 n\times m$

$x_2 = 0.6 m$

Answer: $x_2 = 0.6 m$