A small mouse has gotten into a peculiar situation. Somehow, the mouse has found

Question

A small mouse has gotten into a peculiar situation. Somehow, the mouse has found himself hanging off the end of the metal structure shown in the figure below. Hopefully the structure can support the 37.8 g mouse because it would not make the fall. The structure is rigidly fixed to a wall and has mass per unit length of 5.73 g/cm and dimensions x= 19.8 cm and y = 39.2 cm, defined as y = 0.1x^2 (where the attachment point P has coordinates xp = 0, yp = 0). Calculate the torque about the attachment point P.

Explanation / Answer

d = [(dx)² + (dy)²]

or since y = 0.1x²

dy/dx = 0.2x

therefore d = [(1)² + (dy/dx)²] dx

d = [1 + (0.2x)²]dx

weight = dmg = 5.73dg(in grams) = 5.73g[1 + (0.2x)²]dx

moment about P is x{5.73g[1 + (0.2x)²]dx}

5.73g x[1 + (0.2x)²]dx from x = 0 to x = 19.8 cm

= 5.73 g x [489.82 grams.cm] = 5.73(9.81)(489.82)/(100000) = 0.2753 Nm

=> net clockwise torque = 0.2753 + 0.0378 x (0.198) = 0.2827 Nm

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