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Calculate the force the biceps muscle must exert to hold the forearm and its load as shown in the figure below, 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.


(a) The figure shows the forearm of a person holding a book. The biceps exert a force FB 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

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 FB, that of the elbow joint is FE, that of the weights of the forearm is wa , and its load is wb. Two of these are unknown FB, so that the first condition for equilibrium cannot by itself yield FB . But if we use the second condition and choose the pivot to be at the elbow, then the torque due to FE is zero, and the only unknown becomes FB .

Group

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) becomes

7. Force (Newton's second law)
FForce (N)
mMass (kg)
aAcceleration (m/s2)
Variables...
  valueunitslink
Fwa
m
a
2. Torque
τTorque (N*m)
rLever Arm Length (m)
FForce (N)
Variables...
  valueunitslink
ττ2
rr2
Fwa
Group
8. Force (Newton's second law)
FForce (N)
mMass (kg)
aAcceleration (m/s2)
Variables...
  valueunitslink
Fwb
m
a
4. Torque
τTorque (N*m)
rLever Arm Length (m)
FForce (N)
Variables...
  valueunitslink
ττ3
rr3
Fwb
Group

Note that sin θ = 1 for all forces, since θ = 90º for all forces. This equation can easily be solved for FB in terms of known quantities,yielding. Entering the known values gives

1. Mechanical equilibrium - 3=3 Torque example
M1a torque (N*m)
M2a torque (N*m)
M3a torque (N*m)
M4a torque (N*m)
M5a torque (N*m)
M6a torque (N*m)
Variables...
  valueunitslink
M1
M2
M3
M4
M5
M6
Group

which yields

3. Torque
τTorque (N*m)
rLever Arm Length (m)
FForce (N)
Variables...
  valueunitslink
ττ1
rr1
FFB
12. Addition
xsum (dimensionless)
aaddend (dimensionless)
baddend (dimensionless)
Variables...
  valueunitslink
x
a
b
Group

Now, the combined weight of the arm and its load is known, so that the ratio of the force exerted by the biceps to the total weight is

11. Division
xquotient (dimensionless)
adividend (dimensionless)
bdivisor (dimensionless)
Variables...
  valueunitslink
x
a
b

Discussion

This means that the biceps muscle is exerting a force 7.38 times the weight supported.

Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/

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