Question 7.12: Lifting incorrectly from a bent position Estimate the magnit......
Lifting incorrectly from a bent position
Estimate the magnitude of the force that the back muscle in the woman’s back in Figure 7.24 exerts on her backbone and the force that her backbone exerts on the disks in her lower back when she lifts an 18-kg barbell. The woman’s mass is 55 kg. Model the woman’s upper body as a rigid body.
■ The back muscle attaches two-thirds of the way from the bottom of her l = 0.60@m-long backbone and makes a 12° angle relative to the horizontal backbone.
■ The mass of her upper body is M = 33 kg centered at the middle of the backbone and has uniform mass distribution. The axis of rotation is at the left end of the backbone and represents one of the disks in the lower back.
Sketch and translate The figure below is our mechanical model of a person lifting a barbell. We want to estimate the magnitudes of the force TM on B that the back muscle exerts on the backbone and the force FD on B that the disk in the lower back exerts on the backbone. The force that the disk exerts on the bone is equal in magnitude to the force exerted by the backbone on the disk. The upper body (including the backbone) is the system of interest, but we consider the back muscle to be external to the syste since we want to focus on the force it exerts on the backbone. The hinge where the upper body meets the lower body is the axis of rotation.
Simplify and diagram We next draw a force diagram for the upper body. The gravitational force that Earth exerts on the upper body F_{\mathrm{E} \text { on } \mathrm{B}} at its center of mass is M g=(33 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})=323 \mathrm{~N}(73 \mathrm{lb}) The barbell exerts a force on the upper body equal to m g=(18 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})=176 \mathrm{~N}(40 \mathrm{lb}) . Because of our choice of axis of rotation, the force exerted by the disk on the upper body F_{\mathrm{D} \text { on } \mathrm{B}} will not produce a torque. The gravitational force exerted by Earth on the upper body and the force that the barbell exerts on the upper body have clockwise turning ability, while the tension force exerted by the back muscles on the upper body has counterclockwise turning ability.
Represent mathematically The torque condition of equilibrium for the upper body is
\Sigma \tau=+\left(F_{\text {D on B }}\right)(0)+\left[-(M g)(l / 2) \sin 90^{\circ}\right]+\left(T_{\mathrm{M} \text { on B }}\right)(2 l / 3) \sin 12^{\circ}+\left[-\left(F_{\text {Barb on B }}\right)(l) \sin 90^{\circ}\right]
= 0
The x- and y-component forms of the force condition of equilibrium for the backbone are
\Sigma F_x=F_{\mathrm{D} \text { on } \mathrm{B} x}+\left(-T_{\mathrm{M} \text { on } \mathrm{B}} \cos 12^{\circ}\right)=0
\Sigma F_y=F_{\mathrm{D} \text { on } \mathrm{B} y}+T_{\mathrm{M}\text{ } \mathrm{on}\text{ } \mathrm{B}} \sin 12^{\circ}+(-m g)+(-M g)=0
\text { where } F_{\mathrm{D} \text { on } \mathrm{B} x} \text { and } F_{\mathrm{D} \text { on } \mathrm{B} y} are the scalar components of the force that the disk exerts on the upper body.
Solve and evaluate We can solve the torque equation immediately to determine the magnitude of the force that the back muscle exerts on the backbone:
T_{\mathrm{M} \text { on B }}=\frac{(M g)(l / 2)\left(\sin 90^{\circ}\right)+(m g)(l) \sin 90^{\circ}}{(2 l / 3)\left(\sin 12^{\circ}\right)}Note that the backbone length l in the numerator and denominator of all of the terms in this equation cancels out. Thus,
T_{\mathrm{M} \text { on B }}=\frac{(M \text{g})(1 / 2)\left(\sin 90^{\circ}\right)+(m \text{g})(1) \sin 90^{\circ}}{(2 / 3)\left(\sin 12^{\circ}\right)}=\frac{(323 \mathrm{~N})(0.50)(1.0)+(176 \mathrm{~N})(1)(1.0)}{(0.667 \mathrm{~m})(0.208)}
= 2440 N (550 lb)
We then find F_{\mathrm{D} \text { on } \mathrm{B} x} from the x component force equation:
F_{\mathrm{D} \text { on } \mathrm{B} x}=+T_{\mathrm{M} \text { on } \mathrm{B}} \cos 12^{\circ}= +12440 N2 cos 12°
= +2390 N
and F_{\mathrm{D} \text { on } \mathrm{B} y} from the y component force equation:
F_{\mathrm{D} \text { on } \mathrm{B} y}=+M g+m g-T_{\mathrm{M} \text { on B }} \sin 12^{\circ}= +323 N + 176 N – (2440 N)(sin 12°)
= -8 N
Thus, the magnitude of F_{\mathrm{D} \text { on B }} \text { is }
F_{\mathrm{D} \text { on } \mathrm{B}}=\sqrt{(2390 \mathrm{~N})^2+(-8 \mathrm{~N})^2}=2390 \mathrm{~N}(540 \mathrm{lb})
The direction of \vec{F}_{D on B} be determined using trigonometry:
\tan \theta=\frac{F_{\mathrm{D} \text { on } \mathrm{B} y}}{F_{\mathrm{D} \text { on } \mathrm{B} x}}=\frac{-8 \mathrm{~N}}{2390 \mathrm{~N}}=-0.0033
or θ = 0.19° below the horizontal. We’ve found that the back muscles exert a force more than four times the gravitational force that Earth exerts on the person and that the disks of the lower back are compressed by a comparable force.
Try it yourself: Suppose that a college football lineman stands on top of a 1-inch diameter circular disk. How many 275-lb linemen, one on top of the other, would exert the same compression force on the disk as that exerted on the woman’s disk when she lifts the 40-lb barbell?
Answer: The magnitude of the force exerted on the vertebral disk is equivalent to two linemen (2 × 275 lb = 550 lb) standing on the 1-inch-diameter disk.