Forces and torques acting on the hip joint
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Forces and torques acting on the hip joint
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[ID] => 556910
[post_author] => 12815
[post_date] => 2025-01-09 07:41:11
[post_date_gmt] => 2025-01-09 12:41:11
[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
The gait cycle, or walking cycle, refers the repetitive patterns one makes while ambulating. All of the components of the hip joint shown in Figure 1 are vital to coordinating this process.
Figure 1. Components of the hip joint
There are times during the gait cycle when only one foot is touching the ground, and while in this stance, there are four major forces acting on the leg, as shown in Figure 2: Fm, the force of the gluteal muscles pulling against the greater trochanter of the hip, which is oriented 71 degrees from the horizontal axis; Fr, the reactive force of the acetabulum pressing against the head of the femur which is oriented 77.3 degrees from the horizontal axis; W, the normal force of the floor acting on the bottom of the foot which is equal to a person’s weight; and WL, the weight of the leg which is approximately 0.185 W.
If the point at the upper left corner of the above diagram is used as the pivot point / fulcrum, then the mathematical equations describing the forces and torques acting on the hip joint are as follows:
During movement, the hip joint slides about 3 cm inside the socket with each step. As the hip joint slides, it experiences a frictional force. The frictional force acting on the hip joint is equal to the product of the reactive force of the acetabulum acting on the femoral head and the coefficient of kinetic friction:
Equation 4. Frictional force on the hip joint
Friction is greatly reduced by the synovial fluid which lubricates and fills out the irregularities between the surface of the acetabulum and the surface of the femoral head. The coefficient of kinetic friction of a synovial joint is 0.003. Without the cartilage and synovial fluid to protect from frictional tear, the large forces acting on the hip joint during movement would lead to serious damage.
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[question] => The coefficient of kinetic friction, μk, for dry bone on bone is 0.3. According to the passage, by what factor does the synovial fluid reduce the coefficient of friction?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
The coefficient of kinetic friction, μk, for dry bone on bone is 0.3.
The coefficient of kinetic friction, μk, for a synovial joint is 0.003. This value was provided in the passage.
Divide these two numbers to calculate the answer: 0.3 / 0.003 =100
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[answers] => Array
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[0] => Array
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[each_answer] => A. 1000
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[1] => Array
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[each_answer] => B. 100
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[2] => Array
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[each_answer] => C. 30
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[3] => Array
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[each_answer] => D. 300
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => In terms of the weight, W, what is the vertical component of the reactive force of the acetabulum acting on the femoral head?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
Fr sin 77.3° represents the vertical component of the reactive force of the acetabulum acting on the femoral head.
Analyzed the three equations provided in the passage very carefully to determine which one will allow you solve for Fr sin 77.3° in terms of the weight, W.
Use the torque equation and solve for Fr sin 77.3°
∑ τ = 7 x [Fr sin 77.3° – 2.31W] = 0 ; Fr sin 77.3° – 2.31W = 0; Fr sin 77.3° = 2.31W
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[answers] => Array
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[0] => Array
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[each_answer] => A. 0.815 W
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[1] => Array
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[each_answer] => B. 2.43 W
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[2] => Array
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[each_answer] => C. 1.65 W
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[3] => Array
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[each_answer] => D. 2.31 W
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => If an injury causes the gluteal muscle to become severed from the leg, what will happen to the reactive force of the acetabulum on the head of the femur.
[value] => Array
(
[answer] => 4
[description] => Reason for the Correct Answer:
An injury that severs the gluteal muscle implies that the person will no longer be able to apply a force Fm to the greater trochanter of the hip.
Analyzed the three equations provided in the passage very carefully to determine which one will allow you to predict how Fr will respond when Fm is suddenly decreased to zero.
∑ Fx = Fm cos 71° – Fr cos 77.3° =0; if Fm equals zero then in order to satisfy this equation Fr must also equal zero.
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[answers] => Array
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[0] => Array
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[each_answer] => A. It will decrease to half of its strength
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[1] => Array
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[each_answer] => B. It will increase to twice it strength
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[2] => Array
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[each_answer] => C. It will remain unchanged
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[3] => Array
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[each_answer] => D. It will be reduced to zero
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[3] => Array
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[quiz_unique_key] => 2261298308
[question] => How much frictional work is expended in the hip joint of a 75kg male during each step if Fr is equal to 2.4W?
[value] => Array
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[answer] => 3
[description] => Reason for the Correct Answer:
Ff = Fr x μk = 2.4W x 0.003 = 2.4 x 750N x 0.003 = 5.4 N
d is provided in the passage: 3 cm. Converting this into meters will yield 0.03 meters
Frictional Work = Ff x d= 5.4 N x 0.03 meters = 0.162 J
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[answers] => Array
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[0] => Array
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[each_answer] => A. 16.2 Joules
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[1] => Array
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[each_answer] => B. 162 Joules
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[2] => Array
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[each_answer] => C. 0.162 Joules
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[3] => Array
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[each_answer] => D. 1.62 Joules
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[4] => Array
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[quiz_unique_key] => 574431310
[question] => In order to minimize the amount of force that the hip joint experiences, which side should a walking cane be used in a patient ambulating with an injured hip?
[value] => Array
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[answer] => 1
[description] => Reason for the Correct Answer:
A cane will provide an upward force that will partially oppose the patient’s weight.
To maximize the lever arm of the cane should be placed farthest away from the affected hip joint.
By placing the cane in the opposite side of the injured hip, the cane works through a long lever so that a moderate amount of upward force will greatly reduce pressure of the hip joint.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. On the opposite side of the injured hip to maximize the lever arm of the cane
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[1] => Array
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[each_answer] => B. On the same side of the injured hip to minimize the lever arm of the cane
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[2] => Array
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[each_answer] => C. On the same side of the injured hip to maximize the lever arm of the cane
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[3] => Array
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[each_answer] => D. On the opposite side of the injured hip to minimize the lever arm of the cane
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