Question

A light frictionless pulley is suspended with the help of a light spring. One end $P$ of a light inextensible thread that passes over the pulley is free and the other end is tied to another light spring that is affixed to the ground at its lower end as shown in the figure. Stiffness of both the springs is $k=500$ N/m. The free end $P$ is $h=10$ cm above the ground. What minimum pull at the end $P$ will bring it to the ground?

Answer 1

10 N

Answer 2

Let $k$ be the stiffness of the springs and $h$ be the initial height of point $P$ from the ground. Let $T$ be the tension in the thread. Let $x_u$ be the extension of the upper spring and $x_l$ be the extension of the lower spring.

In equilibrium, for the pulley: $kx_u = 2T$. For the lower spring: $T = kx_l$. Combining these, we get $kx_u = 2(kx_l)$, which implies $x_u = 2x_l$.

Let $y_{pulley}$ be the height of the pulley from the ground and $y_{attach}$ be the height of the attachment point of the lower spring from the ground. The total length of the thread $L$ can be expressed as: $L = (y_{pulley} - y_{attach}) + (y_{pulley} - h) = 2y_{pulley} - y_{attach} - h$.

Let $Y$ be the height of the fixed support of the upper spring, and $L_u, L_l$ be the natural lengths of the upper and lower springs, respectively. $y_{pulley} = Y - (L_u + x_u)$ $y_{attach} = L_l + x_l$ Substituting these into the expression for $L$: $L = 2(Y - L_u - x_u) - (L_l + x_l) - h$ $L = 2Y - 2L_u - 2x_u - L_l - x_l - h$ Using $x_u = 2x_l$: $L = 2Y - 2L_u - 2(2x_l) - L_l - x_l - h = 2Y - 2L_u - L_l - 5x_l - h$.

When a pull $F$ is applied at $P$, the tension becomes $T' = T+F$. The new extensions are $x_u'$ and $x_l'$, and the new height of $P$ is $h'$. The force equations are $kx_u' = 2T'$ and $T' = kx_l'$, so $x_u' = 2x_l'$. The new length of the thread is $L = 2y_{pulley}' - y_{attach}' - h'$, where $y_{pulley}' = Y - L_u - x_u'$ and $y_{attach}' = L_l + x_l'$. $L = 2(Y - L_u - x_u') - (L_l + x_l') - h'$ $L = 2Y - 2L_u - L_l - 5x_l' - h'$.

We want to find the minimum pull $F$ that brings $P$ to the ground, so $h'=0$. $L = 2Y - 2L_u - L_l - 5x_l'$. Equating the initial and final expressions for $L$: $2Y - 2L_u - L_l - 5x_l - h = 2Y - 2L_u - L_l - 5x_l'$ $-5x_l - h = -5x_l'$ $5x_l' = 5x_l + h$ $x_l' - x_l = h/5$.

The pull $F$ is the increase in tension: $F = T' - T = kx_l' - kx_l = k(x_l' - x_l)$. Substituting $x_l' - x_l = h/5$: $F = k(h/5)$.

Given $k = 500$ N/m and $h = 10$ cm $= 0.1$ m. $F = (500 \text{ N/m}) \times (0.1 \text{ m} / 5) = 500 \times 0.02 = 10$ N. The minimum pull required is $10$ N.