Question

A flamingo is able to stand on one leg without using any muscles. It can do this by locking its leg joints so that it acts as a rigid body. (a) If the height of the centre of mass is 110 cm above the ground, and the width of the flamingo's foot is 10 cm. What angle would the Flamingo be able to tilt without falling over? We can use a mass of 3.5 kg for the bird 110cm 10cm Figure 1: Flamingo standing on one leg (b) We assume that the Flamingo's joints are locked and the foot is directly below the centre of mass. Draw a force diagram on the Flamingo and calculate the net force and net torque on the bird (c) Assume now that the Flamingo's leg was not locked, The leg can now be modeled as a single rod 100 cm long. The weight of the flamingo's body is applied downwards at the top of the leg and the leg is 3 degrees from vertical. The weight of the leg is 0.20kg. Draw the forces acting on the Flamingo's leg. What is the net torque on the Leg. What does this tell us about the benefits of the Flamingo's leg-locking mechar

Answer 1

Normal reaction by ground can shift over its foot's width that is between 10cm sigma F_y = 0 N = mg sigma c = 0 about p N times 10/2 = mg times d doublerightarrow d = 10/2 = 5cm Now tan theta = 5/10 theta = 2.6 degree(2.6 degree) if of reduces then normal will shift forward point p \r\n d/100 = tan 3 degree d = 5.24 cm Normal will shift to A to prevent rotation m' = 3.5 kg m = 0.2 kg for our calculation we assume normal shift of x distance from p sigma F = 0 N - m^1g - mg = 0 N = 3.7g = 36.26 Newton sigma c = m^1g times 5.24 times 10^-2 + mg times 5.24/2 times 10^-2 -N times x sigma c = 3.5 times 9.8 times 5.24 times 10^-3 + 0.2 times 9.8 times 5.24 times 10^-2 -Nk sigma c = 1.84867 - 36.26 x