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.
Vanessa? Are you really Tom?
See prev post.
(a) To show that each hip supports half of the upper weight when a person is simply standing upright, we can use the concept of equilibrium. In equilibrium, the sum of the forces acting on an object is zero.
In this case, the forces acting on each hip are the weight of the upper body and the force of friction. Since there is no upward acceleration, the vertical forces must balance for equilibrium to occur.
Let's denote the weight of the upper body as W.
For equilibrium, we have:
Sum of vertical forces = 0
The weight of the upper body is acting downwards, so we can consider it as a force:
Weight of upper body = W
Since each hip supports half of the upper body weight in this case, the force exerted by each hip is W/2.
Therefore, each hip supports half of the upper weight.
(b) When a person is walking, the weight distribution and forces acting on each hip change. The maximum friction force at the hip joint can be calculated using the given coefficients of kinetic friction (µk) range.
Let's denote the upper weight as W, person's weight as 65 kg, and the coefficient of kinetic friction as µk.
The maximum friction force can be calculated as:
Maximum friction force = µk * Normal force
The normal force can be calculated as:
Normal force = Weight of upper body + Person's weight
When walking, one leg is off the ground, so the force on each hip is now equal to the total weight of the upper body and the person's weight.
Therefore, the maximum friction force at the hip joint when walking is:
Maximum friction force = µk * (Weight of upper body + Person's weight)
To calculate the maximum kinetic friction force at the hip joint, we need to use the minimum value of the coefficient of kinetic friction (µk = 0.0050) and the weight of the person, which is given as 65 kg.
Maximum friction force = 0.0050 * (Weight of upper body + Person's weight)
= 0.0050 * (0.65 * Person's weight + Person's weight)
= 0.0050 * (0.65 * 65 + 65)
= 0.0050 * (42.25 + 65)
= 0.0050 * 107.25
= 0.536 N
Therefore, the maximum kinetic friction force at the hip joint when walking is 0.536 N.
(c) In the worst case scenario where the synovial fluid disappears and bone-on-bone contact occurs with a coefficient of kinetic friction of 0.30, we can calculate the greatest friction force for the walking person using the same formula as in part (b):
Maximum friction force = µk * (Weight of upper body + Person's weight)
Using the minimum value of the coefficient of kinetic friction (µk = 0.30) and the given weight of the person (65 kg):
Maximum friction force = 0.30 * (Weight of upper body + Person's weight)
= 0.30 * (0.65 * Person's weight + Person's weight)
= 0.30 * (0.65 * 65 + 65)
= 0.30 * (42.25 + 65)
= 0.30 * 107.25
= 32.18 N
Therefore, the greatest friction force for the walking person in this worst case scenario is 32.18 N.