When humans walk, their weight shifts from one foot to the other. While one foot
ID: 1287176 • Letter: W
Question
When humans walk, their weight shifts from one foot to the other. While one foot is off the ground, their centre of mass must be positioned directly above the other foot, so that their body as a whole will be in static equilibrium. This requires a lot of force from the hip abductor muscles, which is applied at the top of the femur. When these muscles are injured, a cane helps to reduce this force when walking. The cane enables the patient to position the body's centre of mass a distance x from the right foot. Calculate x, assuming that the normal force acting on the cane, FC, equals 1/6 of the body weight, and that the remaining weight is supported entirely by the right leg. The cane is located a distance d1 = 27.30 cm from the person's centre of mass. The length of the leg is L = 0.77 m and the person weighs Wb = 776 N.
The socket of the pelvis (acetabulum) exerts a force FA. The net force FM of a series of abductor muscles is applied to the top of the femur (greater trochanter) at an angle of 70
When humans walk, their weight shifts from one foot to the other. While one foot is off the ground, their centre of mass must be positioned directly above the other foot, so that their body as a whole will be in static equilibrium. This requires a lot of force from the hip abductor muscles, which is applied at the top of the femur. When these muscles are injured, a cane helps to reduce this force when walking. The cane enables the patient to position the body's centre of mass a distance x from the right foot. Calculate x, assuming that the normal force acting on the cane, FC, equals 1/6 of the body weight, and that the remaining weight is supported entirely by the right leg. The cane is located a distance d1 = 27.30 cm from the person's centre of mass. The length of the leg is L = 0.77 m and the person weighs Wb = 776 N. The socket of the pelvis (acetabulum) exerts a force FA. The net force FM of a series of abductor muscles is applied to the top of the femur (greater trochanter) at an angle of 70 A degree with respect to the horizontal, at a distance d2 = 6.3 cm from the acetabulum. The centre of mass is located a horizontal distance d3 = 9.8 cm from the acetabulum. Calculate the magnitude of FM. To simplify the calculation, ignore the weight of the leg. Calculate the x and y components of FA. Ignore the weight of the leg.Explanation / Answer
Sum the moments about the right foot:
?M = 0 = (715N / 6)*(x + 33.7cm) - 715N*x
596N * x = 4016 N
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