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natalie 4:01When a person stands on tiptoe (a strenuous position), the position of the foot is as shown in Figure a. The total gravitational force on the body,Fg, is supported by the forcen exerted by the floor on the toes of one foot. A mechanical model of the situation is shown in Figure b, whereT is the force exerted by the Achilles tendon on the foot andR is the force exerted by the tibia on the foot. Find the values of T, R, and whenFg = n = 570 N. (For , enter the smaller of the two possible values between 0° and 90°.)

natalie 4:01When a person stands on tiptoe (a strenuous position), the position of the foot is as shown in Figure a. The total gravitational force on the body,F class=

Answer :

We have the next equations

Sum the vertical forces

[tex]\sum ^{}_{}F_y=Tcos\theta-Rcos(15)+570=0[/tex]

Sum of the horizontal forces

[tex]\sum ^{}_{}F_x=Rsin(15)-Tsin\theta=0[/tex][tex]R=\frac{T\sin (\theta)}{\sin (15)}[/tex]

Sum of the moments about R

[tex](-570)\cos (\theta)(0.18)+T(0.07)=0[/tex]

We isolate here the T

[tex]T=\frac{570\cos\theta(0.18)}{(0.07)}=1465.71\cos (\theta)[/tex]

We substitute the value of T in the equation with R isolated

[tex]R=\frac{1465.71\cos(\theta)\sin(\theta)}{\sin(15)}[/tex]

Ans we substitute the value of R and T in the sum of forces in y

[tex]1465.71\cos ^2(\theta)-\frac{1465.71\cos(\theta)\sin(\theta)}{\sin(15)}cos(15)+570=0[/tex]

We simplify the equation

[tex]\begin{gathered} \cos ^2(\theta)-3.732(\sin (\theta))(\cos (\theta))+0.388=0 \\ \end{gathered}[/tex]

[tex]\cos ^4(\theta)-(0.8809)\cos ^2(\theta)+0.01013=0[/tex]

[tex]\cos ^2\theta=(0.01165)(0.8693)[/tex][tex]\theta=\cos ^{-1}(\sqrt[]{0.8693})=21.2\text{\degree}[/tex]

Then for T

[tex]T=1465.71\cos (21.2)=1366.52N[/tex]

Then for R

[tex]R=\frac{1465.71\cos (21.2)\sin (21.2)}{\sin (15)}=1909.31N[/tex]

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