Static Equilibrium - AP Physics 1

Part I: Laws of Translational Equilibrium

Vertical Translational Balance

The sum of all upward vertical forces must exactly equal the downward gravitational load:

$$\sum F_y = 0$$

Horizontal Translational Balance

The sum of all leftward horizontal forces must exactly balance rightward horizontal forces:

$$\sum F_x = 0$$

Figure 1.1: Symmetric system hanging weights

The Coordinate System Rotation

When an object is suspended by ropes, the vertical and horizontal tension coordinates must cancel out to maintain dynamic equilibrium:

Symmetric Hanging Box Proof:

$$W - [T\cos\theta + T\cos\theta] = 0$$

$$W - 2T\cos\theta = 0$$

$$T = \frac{W}{2\cos\theta}$$

Part II: Asymmetric Hanging System (Off-Center Knot)

Example Problem Derivation

A 10. kg box is supported by two ropes anchored to a flat ceiling at asymmetric angles of $45^\circ$ and $30^\circ$ relative to the horizontal.

Question: Which rope holds more tension?

Figure 2.1: Asymmetric hanging system schematic. The steeper rope on the left ($45^\circ$) holds a greater vertical load, resulting in more tension.

Calculating $T_1$ and $T_2$

Step 1: Set up Horizontal Force Balance ($\sum F_x = 0$)

Leftward pull equals rightward pull:

$$T_{1x} = T_{2x} \implies T_1\cos(45^\circ) = T_2\cos(30^\circ)$$

$$T_1 = T_2\frac{\cos(30^\circ)}{\cos(45^\circ)} \implies T_1 = 1.22 T_2$$

Step 2: Set up Vertical Force Balance ($\sum F_y = 0$)

Ropes support the total weight of the mass ($W = mg = 10.\text{ kg} \times 10.\text{ m/s}^2 = 100\text{ N}$):

$$T_{1y} + T_{2y} = W \implies T_1\sin(45^\circ) + T_2\sin(30^\circ) = 100\text{ N}$$

Step 3: Substitute and Solve for Tension Components

Substitute $T_1 = 1.22 T_2$ into vertical equation:

$$[1.22 T_2]\sin(45^\circ) + T_2\sin(30^\circ) = 100\text{ N}$$

$$0.86 T_2 + 0.50 T_2 = 100\text{ N} \implies 1.36 T_2 = 100\text{ N}$$

Tension 2

$$T_2 = 74\text{ N}$$

Tension 1

$$T_1 = 90\text{ N}$$