You are working for a manufacturing company, which is mathematically given as
- [tex]m=3\sqrt{2}[/tex]
- [tex]m=\frac{15\sqrt{5}}{16}[/tex]
- x=0.747a
- [tex]m/n=\frac{(x^2+a^2)3/2}{x^3}[/tex]
What is the value of m that will place the movable bead in equilibrium at x-a a ....?
a)
Generally, the equation for the force of equilibrium is mathematically given as
F=2fcos\theta
Therefore
[tex]K(npq^2/a^2)=2\frac{kmpq^2}{a\sqrt{2}^2}0.5\\\\np=mP/ \sqrt{2}\\\\where n=3[/tex]
[tex]m=3\sqrt{2}[/tex]
b)
By force equilibrium
[tex]K(npq^2/(2a^2))=2*\frac{kmpq^2}{a\sqrt{5a}^2}* \frac{2a}{\sart{5a}}[/tex]
Therefore
[tex]n/4=2/5*m*2/\sqrt{5}\\\\m= \frac{5\sqrt{5}}{16}\\\\\\\\[/tex]
[tex]m=\frac{15\sqrt{5}}{16}[/tex]
c)
[tex]K(npq^2/x^2)=2\frac{kmpq^2}{a\sqrt{x^2+a^2}^2}0.5*x/\sqrt{x^2+a^2}[/tex]
x^2+a^2=(14/3)^{2/3}x^2
x=a/1.338
x=0.747a
d)
By force equilibrium
[tex]K(npq^2/x^2)=2\frac{kmpq^2n}{\sqrt{x^2+a^{3/2}}^2}\\\\n/x^2=\frac{2mx}{(x^2+a^2)^{3/2}}[/tex]
[tex]m/n=\frac{(x^2+a^2)3/2}{x^3}[/tex]
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