A person bending forward to lift a load with his back, as shown in the figure, rather than with his knees can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in the figure of a person bending forward to lift a 140 N object. The spine and upper body are represented as a uniform horizontal rod of weight 380 N, pivoted at the base of the spine. The erector spinalis muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12.0°. Find the tension in the back muscle (T).
To find the tension in the back muscle (T), we can use the principle of torque equilibrium. Torque is the rotational force that causes an object to rotate.
Let's break down the forces acting on the system:
1. Weight of the uniform horizontal rod (380 N): This force acts downward and is exerted at the center of mass of the rod. We can consider this force as acting at the pivot point.
2. Weight of the object being lifted (140 N): This force acts downward and is exerted at the center of mass of the object. We can also consider this force as acting at the pivot point.
3. Tension in the back muscle (T): This force acts upward and is exerted at a point two-thirds of the way up the spine. This is the force that we need to find.
To maintain equilibrium, the sum of the torques about the pivot point must be zero. The torque produced by a force is given by the product of the force and the perpendicular distance from the pivot point to the line of action of the force.
Let's denote the perpendicular distance from the pivot point to the line of action of the tension force T as d. Since the erector spinalis muscle is attached two-thirds of the way up the spine, we can calculate this distance.
d = (2/3) * length of the spine
Now, let's calculate the torques:
Torque due to the weight of the uniform horizontal rod = 380 N * 0 = 0 Nm (since the force acts at the pivot point).
Torque due to the weight of the object being lifted = 140 N * 0 = 0 Nm (since the force acts at the pivot point).
Torque due to the tension in the back muscle = T * d
Since the system is in equilibrium, the sum of the torques is zero:
0 Nm + 0 Nm + T * d = 0 Nm
Therefore,
T * d = 0 Nm
Substituting the value of d, we get:
T * (2/3) * length of the spine = 0 Nm
Now, we can solve for T:
T = 0 Nm / [(2/3) * length of the spine]
Please provide the length of the spine, and I will be able to calculate the tension in the back muscle.