Question

On applying a force F, point is displaced vertically down by y from equilibrium position. Find force F in terms of the force constant k of the spring and displacement y, for the cases (a) and (b), as shown in fig.

Answer 1

Case (a)
At point $P:F=T$........(i)
And for the equilibrium of pulley,
$2T=F_s$........(ii)
Spring stretches by (y/2). so
$F_s=k\left(y/2\right)$......(iii)
So substituting $F_s$ from (iii) in (ii) and then T from (ii) in (i), we get
$F=\left(k/4\right)y$..........(A)
Case (b)
As the tension in massless string and spring will be same,
$T=F^{\prime }$.........(i)
For pulley :$F=2{F^{\prime }}_s$........(ii)
Now if the mass M shifts by y, the spring will stretch by 2y (as string is inextensible)
${F^{\prime }}_s=k\left(2y\right)$.....(iii)
So substituting F' from Eq. (ii) in (iii),
$F=\left(4k\right)y$..........(B)
Force of spring does not change instantaneously