AP Physics C: Tension in Wires at Angle θ with Vertical

Answer

The correct option is (B). The tension in each wire is $\frac{Mg}{2\cos\theta}$ because the vertical components of the two tensions must sum to counteract the weight of the mass.

Explanation

Observation: The image shows a mass $M$ suspended by two wires. Each wire forms an angle $\theta$ with the vertical axis, not the horizontal. The system is in static equilibrium, meaning the net force in both the $x$ and $y$ directions is zero.

1. Knowns and Setup

   • Mass $= M$
   • Vertical force $= Mg$ (downward)
   • Angle $= \theta$ with the vertical line
   • Number of wires $= 2$ (each with tension $T$)
   • ⚠️ This step is required on exams: Always define your coordinate system and draw a Free Body Diagram (FBD). Here, the positive $y$-axis is vertical and positive $x$-axis is horizontal.

2. Force Decomposition Since the angle $\theta$ is with the vertical, the vertical component of the tension vector $T$ is adjacent to the angle.

   • $T_y = T \cos\theta$
   • $T_x = T \sin\theta$

3. Applying Newton's Second Law Since the object is held at rest (static equilibrium), the sum of all forces in the $y$-direction must be zero ($\sum F_y = 0$). $\sum F_y = (T \cos\theta) + (T \cos\theta) - Mg = 0$ Sum of vertical components of tension minus the weight equals zero.

4. Solving for Tension ($T$) Rearranging the equation to isolate $T$: $2T \cos\theta = Mg$ Double vertical components equal weight. $T = \frac{Mg}{2\cos\theta}$ Divide by 2 and cosine to isolate tension.

Force Component	Formula
Vertical component per wire	$T \cos\theta$
Total upward force	$2T \cos\theta$
Downward weight	$Mg$

Final Answer

$\boxed{\frac{Mg}{2\cos\theta}}$

Common Mistakes

• Confusing sine and cosine: Students often default to using $\sin\theta$ for the vertical component. Always identify which axis your angle is drawn against. If $\theta$ is with the vertical, the vertical component is always assigned to the $\cos$ function.
• Neglecting the number of wires: Forgetting the factor of $2$ is a common error. Since there are two wires sharing the load equally, the weight $Mg$ is distributed across both vertical tension components: $2 \times T_y = Mg$.