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  4. Forces and torques in muscles

11.6 Forces and torques in muscles and joints

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  • Explain the forces exerted by muscles.
  • State how a bad posture causes back strain.
  • Discuss the benefits of skeletal muscles attached close to joints.
  • Discuss various complexities in the real system of muscles, bones, and joints.

Muscles, bones, and joints are some of the most interesting applications of statics. There are some surprises. Muscles, for example, exert far greater forces than we might think. [link] shows a forearm holding a book and a schematic diagram of an analogous lever system. The schematic is a good approximation for the forearm, which looks more complicated than it is, and we can get some insight into the way typical muscle systems function by analyzing it.

Muscles can only contract, so they occur in pairs. In the arm, the biceps muscle is a flexor—that is, it closes the limb. The triceps muscle is an extensor that opens the limb. This configuration is typical of skeletal muscles, bones, and joints in humans and other vertebrates. Most skeletal muscles exert much larger forces within the body than the limbs apply to the outside world. The reason is clear once we realize that most muscles are attached to bones via tendons close to joints, causing these systems to have mechanical advantages much less than one. Viewing them as simple machines, the input force is much greater than the output force, as seen in [link] .

A forearm of a person holding a physics book is shown. The biceps and triceps muscles of the arm are visible. The elbow joint is the pivot point. The upper part of the arm is vertical and the lower part is horizontal. Biceps muscles are applying a force F B upward. The vertical bone of hand exerts a force F E on the pivot. At the midpoint of the lower part of the hand, the center of gravity of the hand is shown where the weight of the hand acts. The midpoint of the front face of the book is its center of gravity, where its weight acts downward. A free body diagram is also shown and the distances of the three forces F-B, C-G of arm, and C-G of book from the pivot are shown as r one, r two and r three.
(a) The figure shows the forearm of a person holding a book. The biceps exert a force F B to support the weight of the forearm and the book. The triceps are assumed to be relaxed. (b) Here, you can view an approximately equivalent mechanical system with the pivot at the elbow joint as seen in [link] .

Muscles exert bigger forces than you might think

Calculate the force the biceps muscle must exert to hold the forearm and its load as shown in [link] , and compare this force with the weight of the forearm plus its load. You may take the data in the figure to be accurate to three significant figures.

Strategy

There are four forces acting on the forearm and its load (the system of interest). The magnitude of the force of the biceps is F B size 12{F rSub { size 8{B} } } {} ; that of the elbow joint is F E size 12{F rSub { size 8{E} } } {} ; that of the weights of the forearm is w a size 12{w rSub { size 8{a} } } {} , and its load is w b size 12{w rSub { size 8{b} } } {} . Two of these are unknown ( F B size 12{F rSub { size 8{B} } } {} and F E size 12{F rSub { size 8{E} } } {} ), so that the first condition for equilibrium cannot by itself yield F B size 12{F rSub { size 8{B} } } {} . But if we use the second condition and choose the pivot to be at the elbow, then the torque due to F E size 12{F rSub { size 8{E} } } {} is zero, and the only unknown becomes F B size 12{F rSub { size 8{B} } } {} .

Solution

The torques created by the weights are clockwise relative to the pivot, while the torque created by the biceps is counterclockwise; thus, the second condition for equilibrium net τ = 0 size 12{ left ("net "τ rSub { size 8{"cw"} } ="net "τ rSub { size 8{"ccw"} } right )} {} becomes

r 2 w a + r 3 w b = r 1 F B . size 12{r rSub { size 8{2} } w rSub { size 8{a} } +r rSub { size 8{3} } w rSub { size 8{b} } =r rSub { size 8{1} } F rSub { size 8{B} } } {}

Note that sin θ = 1 size 12{"sin"θ=1} {} for all forces, since θ = 90º size 12{θ="90"°} {} for all forces. This equation can easily be solved for F B size 12{F rSub { size 8{B} } } {} in terms of known quantities, yielding

F B = r 2 w a + r 3 w b r 1 . size 12{F rSub { size 8{B} } = { {r rSub { size 8{2} } w rSub { size 8{a} } +r rSub { size 8{3} } w rSub { size 8{b} } } over {r rSub { size 8{1} } } } } {}

Entering the known values gives

F B = 0 . 160 m 2 . 50 kg 9 . 80 m/s 2 + 0 . 380 m 4 . 00 kg 9 . 80 m/s 2 0 . 0400 m size 12{F rSub { size 8{B} } = { { left (0 "." "16"" m" right ) left (2 "." "50"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )+ left (0 "." "38"" m" right ) left (4 "." "00"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )} over {0 "." "04"" m"} } } {}

which yields

F B = 470 N . size 12{F rSub { size 8{B} } ="470"" N"} {}

Now, the combined weight of the arm and its load is 6.50 kg 9.80 m/s 2 = 63.7 N size 12{ left (6 "." "50"`"kg" right ) left (9 "." "80"`"m/s" rSup { size 8{2} } right )="63" "." 7`N} {} , so that the ratio of the force exerted by the biceps to the total weight is

F B w a + w b = 470 63 . 7 = 7 . 38 . size 12{ { {F rSub { size 8{B} } } over {w rSub { size 8{a} } +w rSub { size 8{b} } } } = { {"470"} over {"63" "." 7} } =7 "." "38"} {}

Discussion

This means that the biceps muscle is exerting a force 7.38 times the weight supported.
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Source:  OpenStax, Une: physics for the health professions. OpenStax CNX. Aug 20, 2014 Download for free at http://legacy.cnx.org/content/col11697/1.1
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