Problem 1 A person has legs of length L = 1.1 m. (a) If the maximum angle betwee

Question

Problem 1 A person has legs of length L = 1.1 m. (a) If the maximum angle between the legs during walking is = 50o , what is the person’s step length, A? (Assume that the legs are rigid rods of length L.) (b) If the person takes one step every 1.2 s, what is the period T of the gait cycle? (c) Is the period T in this case longer or shorter than the period for most effortless walking? (d) What is the person’s walking speed?

Problem 2 Assume that a person’s leg can be represented as a L = 1.2 m long shaft of bone with an average cross-sectional area of A = 3 cm2 . What is the amount of shortening of the leg, L, when a person of mass m = 70 kg supports all the body weight on this leg? Take the compressive Young’s Modulus for bone as Y = 1.5x1010 Pa.

Problem 3 We would like to make some rough estimates of the forces and stresses on the driver in car crashes without and with seat belts and air bags. Let us assume that the driver’s mass is m = 70 kg and the car’s speed before the collision is v = 72 km/h. We see that the two main effects of the restraining devices are to (i) increase the distance over which the body is decelerated, and (ii) to distribute the decelerating force over a larger area. Find the average stopping force on the body and the corresponding average stress when (a) There are no belts or bag. From the picture we estimate that the total stopping distance for the body in this situation is s = 10 cm and that the stopping force is distributed over the top of the steering column, which has area A = 100 cm2 . (b) Belts and airbag are used. We estimate that the stopping distance is s = 50 cm, and the stopping force is distributed over the full area of the chest and face, with A = 0.25 m2 . (c) Compare with the breaking stress for bone, Sb 108 Pa. (Note that the force and stress values obtained above are average values. Peak values are likely significantly higher.)

a video that might help:   http://www.youtube.com/watch?v=d7iYZPp2zYY&NR=1

Explanation / Answer

a) step length = 2Lsin25 = 2.2 sin25 = 0.92976 m

b) period = 2 x 1.2 = 2.4 s

c) longer

d) walking speed = 0.92976/1.2 = 0.7748 m/s

problem2

stress = weight/area = mg/A = 70 x 9.8/(0.0003) = 2286666.667 N/m2

strain = stress/Y = 2286666.667/1.5 x 1010

=> L = 1.524444 x 10-4 L = 1.524444 x 10-4 x 1.2 = 1.829333 x 10-4 m = 0.1829333 mm

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