Pregnancy shift of com. The figure below gives a simplified...

Pregnancy shift of com. The figure below gives a simplified view of the weight supported on each foot of someone with weight mg standing upright with half the weight supported by each foot. An x axis lies along the foot, with $x_f = 0$ at the forefoot and $x_h = 25$ cm at the heel. On each forefoot the downward force is $f(mg/2)$, where $f$ is the fraction of the weight on a foot that is supported by the forefoot. Similarly, on each heel the downward force is $h(mg/2)$. The foot (and the person) is in equilibrium around the center of mass. During pregnancy, the center of mass shifts toward the heels by 41 mm. What percent of the weight shifts toward the heels? (The center of mass shifts back to its initial position after pregnancy.)

Transcript

00:01 All right so let here we are given that is equal to 13 centimeters okay and b is equal to 23 .5 centimeters and c is equal to 53 centimeters okay and beta is equal to this is the angle which is 47 degrees and theta is 17 degrees okay and pores at point m is 1940 newton and w01 is 163 newton okay so i have drawn a diagram using a pen and a paper so it is easy okay now you can see this here you can see this okay so this is the x and a y x is and this is ports and this is also ports and this is also a force okay and this is the beta angle and this is also beta here and here okay and this is theta which is 17 degrees okay and this beta is 47 degrees right now a little part a here okay let in part a the moment about point o is equal to the sum of moments about point o is equal to 0 is which is a times epa m sine theta so this is equal to w o 1 cos beta plus b plus w o cause beta times c okay so w not is equal to a sine up theta minus w o 1 b cause of beta divided by c c cata okay now if you put the values here okay so let's put this so this is 1940 times 13 sign up 17 minus 163 times 23 .5 cars up 47 degrees okay divided by times cars are 47 degrees okay so this is equal to 181 .72 new tons okay so let here if we look at part b here so the sum of a pf x is equal to zero so this is if j cars are minus f m cause or plus beta this is equal to 0 and if j cause this is equal to a this is equal to a sub n kata plus beta okay and if j cause this is equal to nineteen hundred forty cause of seventeen plus forty seven so if j cause up this is equal to 150 point four for newton okay let card this equation one now let the sum of forces in the y x is equal to 0 which is minus if j sine up p plus f m sine of theta plus beta minus w one minus w zero is equal to zero and if j sign up this is equal to if m sign up theta plus beta minus w one minus w zero okay so if j sign up this is equal to nineteen hundred forty sign up seventeen plus forty seven minus one hundred sixty three minus one hundred thirty one point seven so finally if j will if j sign the f j sign up this will be equal to fourteen hundred forty eight point two newtons okay now let here i's let square both the equations and add it together okay so this will be equal to if j square times sine square plus car square so this is a trigonometric identity okay which is equal to fourteen hundred 48 .94 square plus 8 .50 .44 square so if j is equal to square root up 1448 .94 square plus 850 .4 square okay so if j is equal to 1680 .08 newtons okay now in part c combining these equations the tangent theta will be equal to 1448 .94 divided by 850 .444.

05:54 So theta will be nothing but tangent inverse of 1448 .94 divided by 850 .44...

Question

Please give Ace some feedback