Do Not Lift with Your Back

By now, we sense that the second condition for equilibrium is a good place to start, and inspection of the known values confirms that it can be used to solve for $F_{\text{B}}–$ if the pivot is chosen to be at the hips. The torques created by ${\mathbf{\text{w}}}_{\text{ub}}$ and ${\mathbf{\text{w}}}_{\text{box}}–$ are clockwise, while that created by ${\mathbf{\text{F}}}_{\text{B}}–$ is counterclockwise.

Using the perpendicular lever arms given in the figure, the second condition for equilibrium $\left(\text{net}\mathbf{\tau }=\text{0}\right)$ becomes

Solving for $F_{\text{B}}$ yields

The ratio of the force the back muscles exert to the weight of the upper body plus its load is

This force is considerably larger than it would be if the load were not present.

More important in terms of its damage potential is the force on the vertebrae ${\mathbf{\text{F}}}_{\text{V}}$. The first condition for equilibrium ($\text{net}\mathbf{\text{F}}=0$) can be used to find its magnitude and direction. Using $y$ for vertical and $x$ for horizontal, the condition for the net external forces along those axes to be zero

Starting with the vertical ($y$) components, this yields

Thus,

yielding

Similarly, for the horizontal ($x$) components,

yielding

The magnitude of ${\mathbf{\text{F}}}_{\text{V}}$ is given by the Pythagorean theorem:

The direction of ${\mathbf{\text{F}}}_{\text{V}}$ is

Note that the ratio of $F_{\text{V}}$ to the weight supported is

This force is about 5.6 times greater than it would be if the person were standing erect. The trouble with the back is not so much that the forces are large—because similar forces are created in our hips, knees, and ankles—but that our spines are relatively weak. Proper lifting, performed with the back erect and using the legs to raise the body and load, creates much smaller forces in the back—in this case, about 5.6 times smaller.