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 each hip supports half of the upper weight when a person is standing upright, we can consider the concept of equilibrium. In equilibrium, the sum of all forces acting on an object is zero.
When standing upright, the person's body is not accelerating vertically, so the net vertical force acting on the body must be zero. This means that the force exerted by each hip joint must balance out the weight of the upper body.
Let's assume the person's total weight is W. Since approximately 65% of the weight is above the hip, the upper weight is given by 0.65W.
Since the person is standing upright, the force exerted by each hip joint must be equal to the upper weight divided by 2, in order to balance out the weight:
Force exerted by each hip = (0.65W) / 2
So, when a person is standing upright, each hip supports half of the upper weight.
(b) When a person is walking, the upper weight shifts from one leg to the other as the person takes steps. During this time, the body comes down fairly hard on each leg. To calculate the maximum kinetic friction force at the hip joint when walking, we need to consider the maximum possible coefficient of kinetic friction and the upper weight.
The maximum possible coefficient of kinetic friction is given as 0.0200. Let's assume the upper weight for the person is still given by 0.65W.
The maximum force of friction can be calculated using the formula:
Maximum force of friction = coefficient of kinetic friction * normal force
In this case, the normal force is the force exerted by each hip joint, which can be calculated as:
Force exerted by each hip = (0.65W) * 2.5
Plugging in the values, we get:
Maximum force of friction = (0.0200) * ((0.65W) * 2.5)
(c) In the worst case scenario, where the synovial fluid disappears, the bones are in direct contact, and the coefficient of kinetic friction increases to 0.30. We need to calculate the greatest friction force for the walking person in this case.
Using the same equation as before, the maximum force of friction can be calculated as:
Maximum force of friction = coefficient of kinetic friction * normal force
In this case, the normal force is still the force exerted by each hip joint, which can be calculated as:
Force exerted by each hip = (0.65W) * 2.5
Plugging in the values, we get:
Maximum force of friction = (0.30) * ((0.65W) * 2.5)
The increased friction in this scenario can cause pain and wear down the joint even more.