Synovial joints. As a person walks, the femur (upper leg bone) slides in a socke
ID: 777560 • Letter: S
Question
Synovial joints. As a person walks, the femur (upper leg bone) slides in a socket in the hip. This socket contains cartilage and synovial fluid, giving it a low coefficient of friction. Consider a 72 kgperson walking on a horizontal sidewalk. Typically, both legs together comprise 34.5% of a person's weight.
Part A: Find the frictional force on the bone in the hip joint. Assume =0.003.
Part B: If this person were walking on the Moon, where g = 1.67 m/s2, what would be the frictional force at the hip joint?
Part C: With age or with osteoarthritis, the synovial fluid can dry up, resulting in a much higher coefficient of friction. If this coefficient increases 100-fold (which can happen), what would be the frictional force (on Earth)?
PartA Find the frictional force on the bone in the hip joint. Assume #0.003. Synovial joints. As a person walks, the femur (upper leg bone) slides in a socket in the hip. This socket contains cartilage and synovial fluid, giving it a low coefficient of friction. Consider a 72 kg person walking on a horizontal sidewalk. Typically, both legs together comprise 34.5% of a person's weight Express your answer to two significant figures and include the appropriate units. F-1 Value Units Submit Incorrect; Try Again; 4 attempts remaining Part B If this person were walking on the Moon, where g 1.67 m/s2, what would be the frictional force at the hip joint? Express your answer to two significant figures and include the appropriate units. Part C With age or with osteoarthritis, the synovial fluid can dry up, resulting in a much higher coefficient of friction. If this coefficient increases 100-fold (which can happen), what would be the frictional force (on Earth)? Express your answer to two significant figures and include the appropriate unitsExplanation / Answer
a) mass of the upper body = 100 - 34.5 %
m = 65.6 % of 72 Kg
m = (65.5/100 )72
m = 47.16 Kg
The weight of the upper body
W = mg = 47.16(9.81)
W = 462.64 N
The normal force applied by the legs on upper body is
N = W = 462.64 N
The frictional force is given by
f = uN
f = 0.003(462.64)
f = 1.4 N
b) if person were walking on moon
g = 1.67 m/s2
W = mg = 47.16(1.67)
W = 78.75 N
The normal force applied by the legs on upper body is
N = W = 78.75 N
The frictional force is given by
f = uN
f = 0.003(78.75)
f = 0.24 N
c) The normal force on Earth is
N = W = 462.64 N
The frictional force is given by
f = uN
u = 100(0.003) = 0.3
f = 0.3(462.64)
f = 138.8 N
in two significant digits
f = 140 N
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