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Transcript
Muscular System notes (Marieb & Hoehn chapter 10)
WCR Oct 2016
Know the information as specified for the muscles in the Muscles to Know list
(…/notes/MusclesToKnow.docx). For the muscles in the list, you should know where they are
in the body and what their actions are. Be able to identify the muscle on a picture. Be able to
translate a muscle’s textbook action into a real-world activity. For example, since you know that
the triceps brachii extend the elbow, then you also understand that the triceps brachii are
important for doing push-ups, since elbow extension is a key part of doing a push-up. For the
bolded muscles, also know their origins and insertions. For bolded muscles with multiple heads
or parts, know the origins and insertions of all heads or parts. If a muscle is not on the list, it will
not be on the test.
We looked at four limb cross sections: two upper limb and two lower limb. It is likely that cross
sectional drawings will appear on the test. Be familiar with those drawings, at the level of detail
that they appear in the textbook.
The test for this chapter will not include innervations (i.e. what nerve innervates what muscle).
The test will not include questions about arrangement of fascicles. It will not include questions
about muscle naming conventions, i.e. no questions about the meaning of word roots such as
albus, brevis, rectus, etc.
____________________________
Know the basic principles of levers, force, distance from the fulcrum, and how that applies to
muscles. We discussed this with examples in class. I summarize the same information below.
The summary takes a lot of space, but it is not that complicated. It will not account for much of
the test. If you are short on time, you may want to study this last.
____________________________
If you know the distance from the joint center to the load and the distance from the joint center to
the point of muscle attachment, then you can calculate muscle force, if you are given the load. (I
assume that “joint center” = “fucrum”.) Alternatively, you can calculate the load, if you are
given the muscle force. The key concept we discussed in class is that of torque, which equals
force times distance*. The torque exerted by the muscle equals the torque exerted by the load, if
the load is not moving. The force exerted by muscle is multiplied by the distance from the joint
center to the muscle attachment point, to get “torque 1”. The force of the load is multiplied by
the distance from the joint center to the load point, to get “torque 2”. If the subject is supporting
the load, the torques will be equal (but opposite).
(The following text uses the multiplication symbol: . It displays correctly on my version of MS Word but I can’t
guarantee it will display correctly for all.)
Example1
Assume the patellar ligament is attached to the tibial tuberosity with an effective distance of 4
cm from the joint center of rotation, and that the quadriceps femoris muscles can pull on this
ligament with 1000 Newtons of force. How much “kick-out” force (in Newtons) can the subject
generate at the distal end of the tibia, 40 cm from the joint center of rotation?
Answer 1
Remember that
(force  distance) at one location = (force  distance) at another location along the tibia.
In this case:
(quads force)  (effective distance from joint center to quads attachment point) = (distal tibial
load)  (distance from joint center to distal tibia).
Now plug in the numbers for this particular problem:
1000 N  4 cm = (distal tibial load)  40 cm.
Solve for the unknown load at the distal tibia:
distal tibial load = 1000 N  4 cm / 40 cm
therefore
distal tibial load = 100 N.
Example 2
Same system as in Example 1, but now the effective distance from the joint center to the tibial
tuberosity is 3 cm, and the distance from the joint center to the distal tibia is 36 cm. The load at
the distal tibia is 90 N. How much force do the quadriceps need to generate?
Answer 2
Remember that
force  distance = force  distance.
In this case,
(quads force)  (effective distance from joint center to quads attachment point) = (distal tibial
load)  (distance from joint center to distal tibia).
Now plug in the numbers for this particular problem:
(quads force)  3 cm = 90 N  36 cm
Solve for the unknown quads force:
quads force = 90 N  36 cm / 3 cm
therefore
quads force = 90 N  12 = 1080 N.
*The equation
torque = force times distance
assumes that the force direction is perpendicular to the line from the fulcrum (joint center) to the point of force
application. In real life, and in more advanced courses, this assumption is not always true, so a more complicated
equation may be needed. But in this course we will make the “perpendicular assumption” and use the simple
equation above.