College Physics

FB (9.42)

wub+wbox=

4200 N

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 vertebraeFV. The first condition for equilibrium (netF= 0) can be used to

find its magnitude and direction. Usingyfor vertical andxfor horizontal, the condition for the net external forces along those axes to be zero

netFy= 0 and netFx= 0. (9.43)

Starting with the vertical (y) components, this yields

FVy–wub–wbox–FBsin 29.0º = 0. (9.44)

Thus,

FVy = wub+wbox+FBsin 29.0º (9.45)

= 833 N +(4200 N)sin 29.0º

yielding

F (9.46)

Vy= 2.87×10

(^3) N.

Similarly, for the horizontal (x) components,

FVx–FBcos 29.0º = 0 (9.47)

yielding

F (9.48)

Vx= 3.67×10

(^3) N.

The magnitude ofFVis given by the Pythagorean theorem:

(9.49)

FV= FV^2 x+FV^2 y= 4.66×10^3 N.

The direction ofFVis

(9.50)

θ= tan– 1

⎛

⎝

FVy

FVx

⎞

⎠

= 38.0º.

Note that the ratio ofFVto the weight supported is

FV (9.51)

wub+wbox=

4660 N

833 N

= 5.59.

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.

310 CHAPTER 9 | STATICS AND TORQUE

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