Statics
Contents
Applications
© The scientific sentence. 2010

Static equilibrium & muscles
In this application, we will apply again the two static equilibrium
formulas, and realize that the forces in muscles can be extremely large.
1. The triceps muscle:
1.1 The forces acting on the forearm:
1. A forearm pulls downward a rope with a force of 50 N. According to Newton's third law,
the rope exerts a force upward (tension) T of 50 N on the forearm.
This tension in the rope acts at a distance
OB = 30  1.2 = 28.8 cm from the pivot.
2. The weight of the forearm considered as a rod of length 30 cm is
mg = 2 x 9.81 = 19.62 N.
This weight acts downward at the centre of mass, that
is at the lever arm equal to OM = 30/2 cm  1.2 cm = 13.8 cm from the pivot.
3. The force F is exerted by the triceps muscle that pulls on the forearm
upward.
This force acts at a distance OA = 1.2 cm from the pivot.
4. The reaction R of the the bones in the upper arm on the
forearm. It is the downward force at the joint O.
1.2 The forearm in static equilibrium:
The related torques for each force are:
τ (T) = + T OB cos 30
τ (mg) =  mg OM cos 30
τ (R) = 0
τ (F) =  F OA cos 30
The system "forearm" is in static equilibrium, that is
the net force is zero and the net torque is zero.
Translational equilibrium:
Σ F = 0
Rotational equilibrium
τ_{net} = Σ τ_{ext} = 0
1. To determine F, we apply
τ_{net} = 0:
T OB cos 30  mg OM cos 30  F OA cos 30 = 0
Or
T OB  mg OM  F OA = 0
Hence
F = (T OB  mg OM)/ OA
(50 x 28.8  2 x 9.8 x 13.8)/ 1.2 =
975 N
F = 975 N
Remark that the angle of inclination θ = 30^{o} is
irrelevant. That could be any angle including the horizontal.
2. To determine R, we apply
Σ (forces) = 0:
Along the yaxis, we have:
+ T  mg  R + F = 0
Then:
R = T + F  mg =
50 + 975  2 x 9.8 = 1005.38 N
R = 1005 N
2. The biceps muscle:
We need to know the biceps tension F required
to Lift a weight.
We choose as a free body the forearm . Therefore
the acting forces are drawn in its related
free body diagram.
The acting forces are:
1. The weight is Mg, M = 5 kg.
This weight has a lever arm OC cos θ1; with
OC = 30 cm.
Its torque related to the pivot is :
τ(weight) =  M g OC cos θ1
2. The forearm makes an angle
θ1 with the horizontal equal to 30^{o}.
It has a mass of 2 kg. Its weight
acts on the center of mass. The corresponding
lever arm is OM cos θ1 with OM = OC/2 = 15 cm.
Its torque related to the pivot is:
τ(forearm) =  m g (OC/2) cos θ1
3. The tension of the muscle in the biceps is F.
It has a lever arm OA cos θ1 from the pivot with OA = 4 cm.
The tension F is projected into its two components:
F_{x} along the xaxis and F_{y} along the yaxis.
1.
The lever arm of F_{x} is OA sin θ1 .
F_{x} = F sin θ2
Its torque related to the pivot is :
τ(F_{x}) =  F_{x} OA sin θ1 =
 F sinθ2 OA sin θ1
2.
The moment arm of F_{y} is OA cos θ1.
F_{y} = F cos θ2
Its torque related to the pivot is:
τ(F_{y}) = + F_{y} OA cos θ1 =
+ F cos θ2 OA cos θ1
4. The reaction R at the pivot O produces no
torque because the pivot is chosen as
the rotation axis.
The reaction R is inclined because its xcomponent
must balance the xcomponent of the tension F.
Its ycomponent must be directed downward to balance,
about the point A, the negative torque produced by the
weight en C and the weight of the forearm on M. Therefore
R must produce an opposing positive torque about the point A.
Applying the rotational equilibrium: Στ = 0, for
the system "forearm", yields:
τ(weight) + τ(forearm) + τ(F)
+ τ(R) = 0
with
τ(F) = τ(F_{x}) + τ(F_{y})
Therefore
 M g OC cos θ1  m g (OC/2) cos θ1
 F sinθ2 OA sin θ1 + F cos θ2 OA cos θ1
+ 0 = 0
Rearranging gives:
M g OC cos θ1 + m g (OC/2) cos θ1 =
F OA (cos θ2 cos θ1  sinθ2 sin θ1)
Using the trigonometric formula:
cos(a + b) = cos a cos b  sin a sin b, gives:
cos θ2 cos θ1  sinθ2 sin θ1
= cos( θ1 + θ2)
Hence:
M g OC cos θ1 + m g (OC/2) cos θ1 =
F OA cos( θ1 + θ2)
Therefore:
F = g OC cos θ1 (M + m/2) /
OA cos( θ1 + θ2)
F = 9.8 x 30 cos (30) x (5 + 2/2)/ 4 x cos (30 + 5)
= 466.23 N
F = 466.23 N
We can find the above result quickly just by
projecting all the three forces over the line OC
of the forearm, and have:
OA F cos (θ1 + θ2)  mg (OC/2) cos θ1  Mg OC cos θ2 = 0
and find as above:
F = g OC (M + m/2) cos θ1 / OA cos(θ1 + θ2)
3. The tension force in the Achilles tendon:
A foot of a person of 60 kg standing on the ball of one foot.
Each foot undergoes a normal N = 60 x 9.8 /2 N. Three
forces act on the foot to maintain this equilibrium:
1. The normal force N,
2. The contact or reaction force R at the ankle joint, and
3. The tension force F in the Achilles tendon.
We want to find the magnitude of (a) the tension F in the
Achilles tendon and the reaction force at the
ankle joint.
At the point O:
Σ τ = 0
That gives:
OA F = OB N
Therefore:
F = (OB/OA) N = (16/4) 60 x 9.8/2 = 1176 N
F = 1176 N
The translational equilibrium is written as:
Σ F = 0
That gives:
N  R + F = 0
Therefore:
R = N + F = + 30 + 1176 = 1206 N
R = 1206 N

