GSO003 Problem 8

1. Force Balance on the Movable Pulley

Let the movable pulley be at $\text{P}(x, y)$P(x,y).

Since the movable pulley is smooth, the tension throughout the string equals $F$F. The pulley is acted upon by three forces:

1. Tension $F$F directed toward $\text{A}$A
2. Tension $F$F directed toward $\text{B}$B
3. Gravity $Mg$Mg downward

Let $\theta_\text{A}$θA​ and $\theta_\text{B}$θB​ be the angles each string segment makes with the vertical.

Horizontal equilibrium:

$$-F\sin\theta_\text{A} + F\sin\theta_\text{B} = 0 \implies \theta_\text{A} = \theta_\text{B} = \theta$$−FsinθA​+FsinθB​=0⟹θA​=θB​=θ

The two string segments always make equal angles with the vertical.

Vertical equilibrium:

$$2F\cos\theta = Mg \quad \cdots(1)$$2Fcosθ=Mg⋯(1)

2. Deriving the Trajectory of the Movable Pulley

Expressing $\theta_\text{A} = \theta_\text{B} = \theta$θA​=θB​=θ in coordinates (with $y < 0$y<0 since $\text{P}$P is below $\text{B}$B):

From $\text{A}(0, H)$A(0,H) to $\text{P}(x, y)$P(x,y): $\tan\theta_\text{A} = \frac{x}{H - y}$tanθA​=H−yx​

From $\text{B}(L, 0)$B(L,0) to $\text{P}(x, y)$P(x,y): $\tan\theta_\text{B} = \frac{L - x}{-y}$tanθB​=−yL−x​

Setting these equal:

$$\frac{x}{H - y} = \frac{L - x}{-y}$$H−yx​=−yL−x​

Rearranging:

$$-xy = (H - y)(L - x)$$−xy=(H−y)(L−x) $$2xy - Ly - Hx + HL = 0$$2xy−Ly−Hx+HL=0 $$y = \frac{H(x - L)}{2x - L} \quad \cdots(2)$$y=2x−LH(x−L)​⋯(2)

This is a rectangular hyperbola. The domain is $\frac{L}{2} < x < L$2L​<x<L.

3. Position and Angle at $x = \frac{3}{4}L$x=43​L

Substituting into equation (2):

$$y = \frac{H\left(\frac{3}{4}L - L\right)}{2 \cdot \frac{3}{4}L - L} = \frac{H \cdot \left(-\frac{1}{4}L\right)}{\frac{1}{2}L} = -\frac{H}{2}$$y=2⋅43​L−LH(43​L−L)​=21​LH⋅(−41​L)​=−2H​

The movable pulley is at $\text{P}\!\left(\frac{3}{4}L,\ -\frac{H}{2}\right)$P(43​L, −2H​).

Computing $\tan\theta$tanθ:

$$\tan\theta = \frac{x}{H - y} = \frac{\frac{3}{4}L}{H + \frac{H}{2}} = \frac{\frac{3}{4}L}{\frac{3}{2}H} = \frac{L}{2H}$$tanθ=H−yx​=H+2H​43​L​=23​H43​L​=2HL​

4. Computing Force $F$F

Substituting $H = 2.0$H=2.0 and $L = 3.0$L=3.0:

$$\tan\theta = \frac{3.0}{2 \times 2.0} = \frac{3}{4}$$tanθ=2×2.03.0​=43​

From the right triangle with $\tan\theta = 3/4$tanθ=3/4: hypotenuse $= \sqrt{3^2 + 4^2} = 5$=32+42​=5, so $\cos\theta = \frac{4}{5}$cosθ=54​.

From equation (1):

$$2F \cdot \frac{4}{5} = Mg \implies F = \frac{5Mg}{8}$$2F⋅54​=Mg⟹F=85Mg​ $$F = \frac{5 \times 16 \times 9.8}{8} = 98\,\text{N}$$F=85×16×9.8​=98N

Answer: 98