Question

An architecture student working at a drafting table requires some muscle action in order to support her head. As shown in the figure, three forces act on her head. If the force provided by muscles isFy = 70 N and the weight of her head is W = 39 N, determine the magnitude and direction of the force provided by the upper vertebrae Fy to hold her head stationary. Assume that this force actsalong a line through the center of mass of her head as do the weight and muscle force. (Assume the +x-axis to be to the right.)

Answer 1

$\begin{gathered} \uparrow+\Sigma Fy=0 \\ Fvy-W-Fmy=0 \\ Fvsin\theta-W-Fmsin(40\text{\degree})=0 \\ Fvs\imaginaryI n\theta=W+Fms\imaginaryI n(40\text{\degree}) \\ W=39N \\ Fm=70N \\ Fvs\imaginaryI n\theta=39N+(70N)s\imaginaryI n(40\text{\degree}) \\ Fvs\mathrm{i}n\theta=84N \\ \rightarrow+\Sigma Fx=0 \\ Fmx-Fvx=0 \\ Fmcos(40\text{\degree})-Fvcos(\theta)=0 \\ Fmcos(40\text{\degree})=Fvcos(\theta) \\ Fvcos(\theta)=Fmcos(40\text{\degree}) \\ Fvcos(\theta)=(70N)cos(40\text{\degree}) \\ Fvcos(\theta)=53.62N \\ Fv^2=(Fvcos(\theta))^2+(Fvs\mathrm{i}n(\theta))^2 \\ Fv=\sqrt{Fvcos(\theta))^2+(Fvsin(\theta))^2} \\ Fv=\sqrt{(53.62N)^2+(84N)^2} \\ Fv=99.7N \\ \theta=\tan^{-1}(\frac{Fvsin(\theta)}{Fvcos(\theta)}) \\ \theta=\tan^{-1}(\frac{84N}{53.62N}) \\ \theta=57.4\text{\degree} \\ The\text{ magnitude is 99.7N} \\ The\text{ angle is 57.4\degree } \end{gathered}$