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Strategy

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 B size 12{F rSub { size 8{B} } } {} if the pivot is chosen to be at the hips. The torques created by w ub size 12{w rSub { size 8{"ub"} } } {} and w box size 12{w rSub { size 8{"box"} } } {} are clockwise, while that created by F B size 12{F rSub { size 8{B} } } {} is counterclockwise.

Solution for (a)

Using the perpendicular lever arms given in the figure, the second condition for equilibrium net τ = 0 size 12{ left ("net "τ rSub { size 8{"cw"} } =" net"τ rSub { size 8{"ccw"} } right )} {} becomes

0 . 350 m 55.0 kg 9.80 m/s 2 + 0.500 m 30.0 kg 9.80 m/s 2 = 0.0800 m F B . size 12{ left (0 "." "35"" m" right ) left ("55" "." 0" kg" right ) left (9 "." "80" m/s" rSup { size 8{2} } right )+ left (0 "." "50"" m" right ) left ("30" "." 0" kg" right ) left (9 "." "80" m/s" rSup { size 8{2} } right )= left (0 "." "08" m" right )F rSub { size 8{B} } } {}

Solving for F B size 12{F rSub { size 8{B} } } {} yields

F B = 4.20 × 10 3 N . size 12{F rSub { size 8{B} } ="4200"" N"} {}

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

F B w ub + w box = 4200 N 833 N = 5.04 . size 12{ { {F rSub { size 8{B} } } over {w rSub { size 8{ ital "ub"} } +w rSub { size 8{"box"} } } } = { {"4200"" N"} over {"833 N"} } =5 "." "04"} {}

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

Solution for (b)

More important in terms of its damage potential is the force on the vertebrae F V size 12{F rSub { size 8{V} } } {} . The first condition for equilibrium ( net F = 0 size 12{"net"`F=0} {} ) can be used to find its magnitude and direction. Using y size 12{y} {} for vertical and x size 12{x} {} for horizontal, the condition for the net external forces along those axes to be zero

net F y = 0 and net F x = 0 . size 12{"net "F rSub { size 8{y} } =0``"and"`"net "F rSub { size 8{x} } =0} {}

Starting with the vertical ( y size 12{y} {} ) components, this yields

F V y w ub w box F B sin 29.0º = 0. size 12{F rSub { size 8{"Vy"} } - w rSub { size 8{"ub"} } - w rSub { size 8{"box"} } - F rSub { size 8{B} } "sin""29" "." 0"°="0} {}

Thus,

F V y = w ub + w box + F B sin 29.0º = 833 N + 4200 N sin 29.0º alignl { stack { size 12{F rSub { size 8{"Vy"} } =w rSub { size 8{"ub"} } +w rSub { size 8{"box"} } +F rSub { size 8{B} } "sin"" 29" "." 0°} {} #" "=" 833 N"+ left ("4200"" N" right )"sin"" 29" "." 0° {} } } {}

yielding

F V y = 2.87 × 10 3 N . size 12{F rSub { size 8{"Vy"} } ="2870"" N"} {}

Similarly, for the horizontal ( x size 12{x} {} ) components,

F V x F B cos 29.0º = 0 size 12{F rSub { size 8{"Vx"} } - F rSub { size 8{B} } "cos"" 29" "." 0"°="0} {}

yielding

F V x = 3.67 × 10 3 N . size 12{F rSub { size 8{"Vx"} } ="3670"" N"} {}

The magnitude of F V size 12{F rSub { size 8{v} } } {} is given by the Pythagorean theorem:

F V = F V x 2 + F V y 2 = 4.66 × 10 3 N. size 12{F rSub { size 8{v} } = left (F rSub { size 8{ ital "Vy"} } rSup { size 8{2} } +F rSub { size 8{ ital "Vx"} } rSup { size 8{2} } right ) rSup { size 8{1/2} } ="4660"" N" "." } {}

The direction of F V size 12{F rSub { size 8{v} } } {} is

θ = tan 1 F V y F V x = 38.0º . size 12{θ="tan" rSup { size 8{ - 1} } left ( { {F rSub { size 8{"Vy"} } } over {F rSub { size 8{"Vx"} } } } right )="38" "." 0°} {}

Note that the ratio of F V size 12{F rSub { size 8{v} } } {} to the weight supported is

F V w ub + w box = 4660 N 833 N = 5 . 59 . size 12{ { {F rSub { size 8{V} } } over {w rSub { size 8{"ub"} } +w rSub { size 8{"box"} } } } = { {"4660"" N"} over {"833 N"} } =5 "." "60"} {}

Discussion

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.

A man is bending forward to lift a box. The back muscles and vertebrae of the person are shown. The weight of the box is acting downward at its center of gravity. The vertebrae of the man are inclined vertical at sixty one degrees. A point on the joint of legs to the upper body is the pivot point. The distance between the center of gravity of the box and the pivot is fifty centimeters and perpendicular distance between the pivot and the weight of the man is thirty five centimeters.
This figure shows that large forces are exerted by the back muscles and experienced in the vertebrae when a person lifts with their back, since these muscles have small effective perpendicular lever arms. The data shown here are analyzed in the preceding example, [link] .

What are the benefits of having most skeletal muscles attached so close to joints? One advantage is speed because small muscle contractions can produce large movements of limbs in a short period of time. Other advantages are flexibility and agility, made possible by the large numbers of joints and the ranges over which they function. For example, it is difficult to imagine a system with biceps muscles attached at the wrist that would be capable of the broad range of movement we vertebrates possess.

There are some interesting complexities in real systems of muscles, bones, and joints. For instance, the pivot point in many joints changes location as the joint is flexed, so that the perpendicular lever arms and the mechanical advantage of the system change, too. Thus the force the biceps muscle must exert to hold up a book varies as the forearm is flexed. Similar mechanisms operate in the legs, which explain, for example, why there is less leg strain when a bicycle seat is set at the proper height. The methods employed in this section give a reasonable description of real systems provided enough is known about the dimensions of the system. There are many other interesting examples of force and torque in the body—a few of these are the subject of end-of-chapter problems.

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Source:  OpenStax, College physics: physics of california. OpenStax CNX. Sep 30, 2013 Download for free at http://legacy.cnx.org/content/col11577/1.1
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