Solution 13
The answer is not correct. We can get it only if the force applied is gradually increased (It would not be the minimum pulling force which would act suddenly). Let the stiffness of both springs be $k$. Let the displacement of the top spring be $x_1$ and the lower spring be $x_2$.
When a force $F$ is applied at the free end $P$, the tension in the thread is $T = F$. This tension $F$ pulls the lower spring and acts on the pulley. The net downward force on the pulley is $2F$, which is balanced by the top spring.
Force balance equations:
$$kx_1 = 2F \implies x_1 = \frac{2F}{k}$$ $$kx_2 = F \implies x_2 = \frac{F}{k}$$The total downward displacement of the point $P$ consists of:
- The downward movement of the pulley: $x_1$.
- The elongation of the lower spring: $x_2$.
- The movement of the thread over the pulley due to pulley displacement: $x_1$ (since the other end of the thread is effectively lowered by $x_1$ by the pulley moving).
Total displacement $\Delta y = 2x_1 + x_2$. For point $P$ to touch the ground, $\Delta y = h$.
$$2\left(\frac{2F}{k}\right) + \frac{F}{k} = h \implies \frac{5F}{k} = h \implies F = \frac{kh}{5}$$Substituting numerical values $k = 500 \text{ N/m}$ and $h = 0.1 \text{ m}$:
$$F = \frac{500 \times 0.1}{5} = \frac{50}{5} = 10.0 \text{ N}$$