First Condition of Equilibrium For translational equilibrium, the vector sum of all external forces must be zero: $\Sigma \vec{F} = 0$. This implies: Sum of forces along x-axis is zero: $\Sigma F_x = 0$ Sum of forces along y-axis is zero: $\Sigma F_y = 0$ Problem Example: Ball in Equilibrium Given Data: Weight of the ball ($w$): $5 \text{ N}$ (downwards) Angle of the rope with the horizontal: $\theta = 60^{\circ}$ Required: Magnitude of horizontal force ($F$) Magnitude of tension in the rope ($T$) Forces Acting on the Ball: Weight ($w$): Acts in the $-y$ direction. Horizontal Force ($F$): Acts in the $+x$ direction. Tension ($T$): Acts upwards and leftwards at $60^{\circ}$ to the horizontal. Resolving Tension ($T$) into Components: Horizontal component ($T_x$): $T_x = T \cos \theta = T \cos 60^{\circ}$ Vertical component ($T_y$): $T_y = T \sin \theta = T \sin 60^{\circ}$ Step 1: Apply $\Sigma F_y = 0$ (Vertical Equilibrium) Upward forces balance downward forces: $$T_y - w = 0$$ $$T \sin 60^{\circ} = w$$ Substitute $w = 5 \text{ N}$: $$T = \frac{w}{\sin 60^{\circ}} = \frac{5 \text{ N}}{0.866}$$ $$\mathbf{T \approx 5.8 \text{ N}}$$ Step 2: Apply $\Sigma F_x = 0$ (Horizontal Equilibrium) Rightward forces balance leftward forces: $$F - T_x = 0$$ $$F = T \cos 60^{\circ}$$ Substitute $T \approx 5.773 \text{ N}$ (from previous step) and $\cos 60^{\circ} = 0.5$: $$F = (5.773 \text{ N}) \times (0.5)$$ $$\mathbf{F \approx 2.9 \text{ N}}$$ Result: Magnitude of horizontal force $F \approx 2.9 \text{ N}$ Magnitude of tension $T \approx 5.8 \text{ N}$
