# What is the magnitude of the gravitational torque about his ankles?

When you stand quietly, you pivot back and forth a very small amount about your ankles. It's easier to maintain stability if you tip slightly forward, so that your center of gravity is slightly in front of your ankles. The 74 kg man shown in (Figure 1) is 1.8 m tall; his center of gravity is 1.0 m above the ground and, during quiet standing, is 5.5 cm in front of his ankles.

What is the magnitude of the gravitational torque about his ankles?

This is the picture

### 1 Answer

- oldschoolLv 72 years agoFavorite Answer
m*g = 726N

Θ = the angle with the vertical of his center of gravity to his ankles.

Θ = atan(1/0.055) = 86.85°

The component of his weight at 90° to the line from his c.o.g. to his ankles = mgcos86.85 = F = 74*9.8*cos86.85 = 39.88N

sin86.85 = 1/L -----> L = 1.0015

39.88*1.0015 = 39.94Nm or about 40Nm

OR

F = mgsin3.15 = 39.85

L*sin86.85 = 1 ----> L = 1.0015

39.85*1.0015 = 39.9 or about 40Nm torque about the ankles.