Question

A traffic light of mass $10\sqrt{3}$ kg is suspended by two cables making $30^{\circ}$ with the vertical. The tension in each cable is ________.

• A. 10 N
• B. 9.8 N
• C. 98 N
• D. 19.6 N
• E. 20 N

Hint:

For symmetrical suspension, $2T\cos\theta = \text{Weight}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
This is a static equilibrium problem. The traffic light is not accelerating, so the net force acting on it is zero. We need to resolve the forces into horizontal and vertical components and apply Newton's First Law (\(\Sigma F = 0\)).
Step 2: Key Formula or Approach
1. Draw a free-body diagram of the traffic light, showing all forces acting on it: its weight acting downwards and the tensions from the two cables acting upwards and outwards.
2. Set up the equilibrium equations: - Sum of horizontal forces is zero (\(\Sigma F_x = 0\)). - Sum of vertical forces is zero (\(\Sigma F_y = 0\)).
3. Solve the equations for the unknown tension.
Step 3: Detailed Explanation
1. Free-Body Diagram and Forces.
- Weight (W): Acts vertically downwards. \(W = mg\). \(m = 10\sqrt{3}\) kg. Let's use \(g \approx 9.8 \text{ m/s}^2\). \(W = 10\sqrt{3} \times 9.8\). - Tensions (T): There are two cables. By symmetry, the tension \(T\) in each cable is the same. Each cable makes an angle of \(30^\circ\) with the vertical.
2. Resolve forces into components.
Let \(T\) be the tension in each cable. - The angle with the vertical is \(30^\circ\). - The vertical component of tension from one cable is \(T \cos(30^\circ)\). - The horizontal component of tension from one cable is \(T \sin(30^\circ)\).
3. Apply Equilibrium Conditions.
- Horizontal forces (\(\Sigma F_x = 0\)): The horizontal component from the left cable is \(-T \sin(30^\circ)\) and from the right cable is \(+T \sin(30^\circ)\). \(-T \sin(30^\circ) + T \sin(30^\circ) = 0\). This equation is automatically satisfied and confirms that the tensions are equal.
- Vertical forces (\(\Sigma F_y = 0\)): The upward forces are the vertical components of the two tensions. The downward force is the weight. (Vertical component from left cable) + (Vertical component from right cable) - (Weight) = 0 \[ T \cos(30^\circ) + T \cos(30^\circ) - W = 0 \] \[ 2T \cos(30^\circ) = W \] 4. Solve for T.
\[ 2T \cos(30^\circ) = mg \] We know \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\).
\[ 2T \left(\frac{\sqrt{3}}{2}\right) = (10\sqrt{3})g \] \[ T\sqrt{3} = 10\sqrt{3}g \] Divide both sides by \(\sqrt{3}\):
\[ T = 10g \] Now, substitute \(g \approx 9.8 \text{ m/s}^2\):
\[ T = 10 \times 9.8 = 98 \text{ N} \] Step 4: Final Answer
The tension in each cable is 98 N.