Mechanics

Contents

Applications

# 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

τnet = Σ τext = 0

1. To determine F, we apply τnet = 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 = 30o is irrelevant. That could be any angle including the horizontal.

2. To determine R, we apply Σ (forces) = 0:

Along the y-axis, 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 30o.
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:
Fx along the x-axis and Fy along the y-axis.

1. The lever arm of Fx is OA sin θ1 .
Fx = F sin θ2
Its torque related to the pivot is :
τ(Fx) = - Fx OA sin θ1 = - F sinθ2 OA sin θ1

2. The moment arm of Fy is OA cos θ1.
Fy = F cos θ2
Its torque related to the pivot is:
τ(Fy) = + Fy 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 x-component must balance the x-component of the tension F. Its y-component 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(Fx) + τ(Fy)

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

 chimie labs | Physics and Measurements | Probability & Statistics | Combinatorics - Probability | Chimie | Optics | contact |