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 160 N object. The spine and upper body are represented as a uniform horizontal rod of weight 420 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 will use the principle of torque equilibrium at the pivot point (base of the spine).
Torque is the product of force and the perpendicular distance from the point of rotation. In this case, the torque due to the weight of the upper body is balanced by the torque due to the tension in the back muscle.
Given:
Weight of the object (W) = 160 N
Weight of the upper body (U) = 420 N
Angle between the spine and the muscle (θ) = 12.0°
First, we need to find the perpendicular distance from the pivot point to the line of action of the weight of the upper body. Let's call this distance 'd'.
Using trigonometry, we can find d:
sin(θ) = d / (2/3 of the length of the spine)
d = (2/3) * L * sin(θ)
Now, we can calculate the torque due to the weight of the upper body:
Torque due to U = U * d
Similarly, we can calculate the torque due to the weight of the object:
Torque due to W = W * (length of the spine)
Since the system is in equilibrium, the torque due to U must be equal to the torque due to W:
Torque due to U = Torque due to W
Therefore, we can write the equation as:
U * d = W * (length of the spine)
Now, we can substitute the given values and solve for the length of the spine:
420 N * [(2/3) * L * sin(12.0°)] = 160 N * L
Simplifying the equation:
L * sin(12.0°) = (160 N * L) / (420 N * (2/3))
Canceling units and rearranging the equation:
L * sin(12.0°) = (8/21) * L
Dividing both sides of the equation by L:
sin(12.0°) = 8/21
Now, we can take the arcsin of both sides to solve for the angle (12.0°):
12.0° = arcsin(8/21)
Using a calculator:
12.0° = 25.59° (approximately)
So, the angle (12.0°) between the spine and the muscle is approximately 25.59°.
Note: At this point, it seems there might be a typographical error or inconsistency in the given information, as the calculated angle is different from the one provided. Please double-check the values given in the problem to ensure accuracy.
Unfortunately, without the correct angle between the spine and the muscle, it is not possible to determine the tension in the back muscle (T).
To solve for the tension in the back muscle, we can use the principle of torque equilibrium, which states that the sum of the torques about any point in a system must be zero.
In this case, we can consider the torque about the pivot point at the base of the spine. The torque exerted by the weight of the upper body can be calculated as follows:
Torque1 = W1 * d1
Where W1 is the weight of the upper body (420 N) and d1 is the distance from the pivot to the center of mass of the upper body.
Similarly, the torque exerted by the tension in the back muscle can be calculated as:
Torque2 = T * d2
Where T is the tension in the back muscle (unknown) and d2 is the distance from the pivot to the point where the back muscle is attached (two-thirds of the way up the spine).
Since the system is in torque equilibrium, the sum of the torques must be zero:
Torque1 + Torque2 = 0
Substituting in the given values:
(W1 * d1) + (T * d2) = 0
We can also express the weight of the upper body in terms of the force exerted by the spine:
W1 = W2 + F
Where W2 is the weight of the 160 N object and F is the force exerted by the spine.
Substituting this into the equation:
((W2 + F) * d1) + (T * d2) = 0
Now, we can solve for the tension in the back muscle (T):
T * d2 = -((W2 + F) * d1)
T = -((W2 + F) * d1) / d2
Substituting the given values:
W2 = 160 N (weight of the object)
d1 = unknown (distance from the pivot to the center of mass of the upper body)
d2 = unknown (distance from the pivot to the point where the back muscle is attached)
F = unknown (force exerted by the spine)
Unfortunately, without the values for d1, d2, and F, we cannot calculate the exact tension in the back muscle (T).
I do not have your picture but if it is a cantilever beam with a cable making a 12 degree angle with the boom 2/3 of the way out we can make some progress.
first what force up is required 2/3 of the way out? Lets call the back 10 units long so there is a 420 N force down at 5 units from the base and a 160 N force down at the tip.
the moments around the base are 420*5+160*10
= 2270 Nm
that must be balanced by the upward component of the muscle tension T
T sin 12 * 6.67 units = 2270
so the tension T = 1637 N or about ten times the weight we are lifting
Now of course we alsoget a compression in the boom(spine) equal to T cos 12 = 1601 N