Consider the hip joint. The bones are normally not in direct contact , but instead are covered with
cartilage to reduce friction. The space between them is filled with synovial fluid, which reduces the
friction further. Due to this fluid, the coefficient of kinetic friction between the bones can range
from 0.00500 to 0.0200. (The wide range of values is due to the fact that motion such as running
causes more fluid to squirt between the bones, thereby reducing friction when they strike each other.)
Approximately 65% of a perosn’s weight is above the hip - we will call this the upper weight.
(a) Show that when a person is simply standing upright, each hip supports half of the upper weight.
(b) When a person is walking, each hip now supports up to 2.5 times the upper weight, depending
on how fast the person is walking. (Recall that when you walk, your weight shifts from one leg
to the other and your body comes down fairly hard on each leg.) For a 65 kg person, what is the
maximum kinetic friction force at the hip joint if µk has a minimum value of 0.0050?
(c) As a person gets older, the aging process, as well as osteoarthritis, can alter the composition of
the synovial fluid. In the worst case, this fluid could disappear, leaving bone-on-bone contact with
a coefficient of kinetic friction of 0.30. What would be the greatest friction force for the walking
person in part (b)? The increased friction causes pain and, in addition, wears down the joint
even more
(a) To show that when a person is simply standing upright, each hip supports half of the upper weight, we can use the concept of equilibrium.
Equilibrium means that the net force and net torque acting on an object are both zero. In this case, we will assume that the person's hip joint is at rest (not moving), which means that the net force acting on the hip joint is zero.
Since the person is not accelerating vertically (assuming constant velocity while standing), the vertical forces must balance out. We can write the equation for equilibrium as:
ΣFy = 0,
where ΣFy represents the sum of the vertical forces.
In standing position, we have two forces acting on the hip joint: the weight of the upper body and the force exerted by the hip joint. We can assume that the only forces acting in the vertical direction are the weight of the upper body (W) and the force exerted by the hip joint (F_hip). Since the person is standing upright, these forces will balance each other:
W = F_hip.
Considering that approximately 65% of a person's weight is above the hip, we can determine the force exerted by the upper body as:
Upper weight = 65% of weight = 0.65 * W.
Each hip supports half of the upper weight, so the force exerted by one hip joint is:
F_hip = (1/2) * Upper weight = (1/2) * 0.65 * W = 0.325 * W.
Therefore, each hip supports half of the upper weight when a person is simply standing upright.
(b) To determine the maximum kinetic friction force at the hip joint when a person is walking, we need to consider the coefficient of kinetic friction (μk) and the upper weight.
We know that the range of μk for the hip joint during motion such as running is 0.00500 to 0.0200. In this case, let's assume the minimum value for μk, which is 0.00500.
The maximum force of kinetic friction (F_friction) can be calculated using the formula:
F_friction = μk * N,
where N is the normal force.
In this case, the normal force is equal to the force exerted by the hip joint, which we established in part (a) as half of the upper weight:
N = 0.325 * W.
Therefore, the maximum kinetic friction force at the hip joint is:
F_friction = 0.00500 * (0.325 * W) = 0.001625 * W.
For a 65 kg person, we can calculate the weight (W) using the formula:
W = mass * acceleration due to gravity = 65 kg * 9.8 m/s^2 = 637 N.
Substituting the value of W into the equation, we get:
F_friction = 0.001625 * 637 N ≈ 1.035 N.
Therefore, the maximum kinetic friction force at the hip joint, when μk has a minimum value of 0.0050, is approximately 1.035 N.
(c) In the worst case scenario, if the synovial fluid completely disappears and there is bone-on-bone contact, the coefficient of kinetic friction (μk) between the bones becomes 0.30.
Using the same formula as in part (b), the maximum kinetic friction force (F_friction) at the hip joint can be calculated as:
F_friction = μk * N,
where N is the normal force.
In this case, the normal force is again equal to half of the upper weight:
N = 0.325 * W.
Substituting the value of μk, we get:
F_friction = 0.30 * (0.325 * W) = 0.0975 * W.
For the weight (W) of a 65 kg person (which we calculated in part (b) as 637 N), the maximum kinetic friction force at the hip joint, with a coefficient of kinetic friction of 0.30, is:
F_friction = 0.0975 * 637 N ≈ 62 N.
Therefore, the greatest friction force for the walking person, in the worst case scenario when the synovial fluid disappears, is approximately 62 N.