Physics 253 In-Class Worksheet
02 Dec 2013
Contact Forces, or “Arthritis, here we come!”
A person holds their arm as shown in the diagram below. The biceps muscle is
attached to the radius bone 2.5 cm from the elbow joint, making an angle of 80◦ to
the horizontal. The entire forearm is 30 cm in length, and has a mass of 2.5 kg.
1. Determine the tension in the biceps.
The tension in the biceps is determined by the counter-torque required to prevent the forearm from rotating under the pull of gravity. That is, the gravitational torque on the forearm is τgrav = M grcm = (2.5)(9.8)(0.15) = 3.68 N · m.
The countertorque is be τbiceps = Ty rb = T sin(80)rb , where rb = 0.025 m is the
distance from the elbow (pivot!) at which the force Ty = T sin(80) is applied.
These torques must be equal in magnitude, so we find
τbiceps = τgrav
=⇒
T =
3.68
M grcm
=
= 149.3 N
rb sin(80)
(0.025) sin(80)
2. Determine the contact forces at the elbow (both magnitude and direction!).
The contact forces are those acting at the elbow that prevent rotation about
the biceps contact. We immediately know one of the contact forces. The
x-component of the tension must be balanced by the Fcx contact force, so
Fcx = T cos(80) = 26 N . As for the Fcy component, this is the force that
must provide a counter-torque to the rotation caused if the pivot were the
biceps contact point. In this case, the center of mass of the arm is a distance r = 0.15 − 0.025 = 0.125 m from the pivot, so taugrav = mgr =
(2.5)(9.8)(0.125) = 3.06 N · m. This will cause the forearm to rotate counterclockwise, so the countertorque must be clockwise. Thus, the contact force must
3.06
= 0.025
= 122.4 N .
act down to prevent rotation, and has a value Fcy = mgr
rb
3. Suppose a 10 lb weight is placed in their hand (basically at the end of the
forearm). Re-calculate the contact forces. What does this tell you about the
wear-n-tear on your joints when lifting weights? [Hint: Ouch!]
With the addition of the weight (m = 4.5 kg) at a distance R = 0.3 m from the
elbow, the net torque equation becomes
τnet = 0 = τbiceps − τarm − τweight
which gives
Ty rb = M g
R
+ mgR
2
=⇒
Ty =
M gR/2 + mgR
rb
Solving for the vertical tension component, we find
Ty =
(2.5)(9.8)(0.15) + (4.5)(9.8)(0.3)
= 676.2 N
0.025
The tension in the biceps is thus T =
Ty
sin(80)
= 686 N .
As in the first part of this worksheet, the contact forces must provide a countertorque to rotation about the biceps contact point. Again, the horizontal force is
completely determined by the biceps tension, Fcx = Tx = T cos(80) = 119 N ,
and this must act to the right . The net force in the vertical direction must be
Fnet;y = 0 = Ty − M g − mg − Fc;y
since we can reason that if the elbow weren’t present, the forearm bone would
rotate clockwise about the biceps contact point (yuk!). This means the contact
force must act down. Solving, we find
Fc;y = Ty − (m + M )g = 606 N
This is obviously much more (a factor of 5 higher) than the contact force without
the weight.
Note that we can also calculate the net torque another way. When the weight
is placed in the hand, the centre of mass of the arm+weight shifts! Since more
weight is further away from the elbow joint, the centre of mass distance increases from rcm = 0.15 m. With a weight in the hand (distance R = 0.30 m
from the elbow), the new center of mass of the arm+weight is located a distance
= 0.25 m from the elbow. The new gravitational torque
rcm = (0.15)(2.5)+(0.3)(4.5)
2.5+4.5
is thus τgrav = (m + M )grcm = (7.0)(9.8)(0.25) = 17.2 N · m in a clockwise direction. Again, the biceps must provide the force to generate the countertorque
τbiceps = rb Ty = rb T sin(80), so it must pull up. Equating the torque magnitudes
gives a tension T =
we found before!
(m+M )grcm
rb sin(80)
=
17.2
(0.025) sin(80)
= 686 N , which is the same thing