Question

. A 200 kg block is attached a the lever AO as shown in the figure. If the stiffiess of the spring is 50 kN/m and the spring is at its natural length when 6-0 find the position (or positions) of equilibrium. 1 = 200 mm m = 200 kg r75 mm k = 50 kN/m-' , Note: in problems l and 2 take the gravity as g=9.81 m/s pointing down

Answer 1

First draw the free body diagram of the given system,

  

= 200 mm 75 mm Kx

Where, T = mg , x = distance moved by spring, Kx = is spring force

  $\sum F_{y} = 0$

  

$-T=Kx$

$T=-Kx$

  $mg=-Kx$

$(200\times 9.81)=-(50000)x$

$1962=-(50000)x$

$x=\frac{1962}{-50000}$

$x=-0.03924m$

$\mathbf{x=0.03924m}$ (spring extended upward)

And

$\sum M_{O} = 0$ , Clock Wise taken as +ve

  $-T\times (lsin\theta )+(Kx)\times r=0$

  $T\times (lsin\theta )=(Kx)\times r$

$1962\times (0.2sin\theta )=(50,000x)\times 0.075$

Substitute x value,

  $1962\times (0.2sin\theta )=(50,000\times 0.03924)\times 0.075$

$1962\times (0.2sin\theta )=147.15$

$sin\theta =\frac{147.15}{392.4}$

  $sin\theta =0.375$

  $\theta =sin^{-1}(0.375)$

  $\mathbf{\theta =22.024^{\circ}}$

So equilibrium positions:

  $\mathbf{x=0.03924m}$ (spring extended upward)

$\mathbf{\theta =22.024^{\circ}}$