What is thyristor family?

**1. Horizontal Force Calculation:**

When the chandelier is displaced horizontally by \( x = 0.17 \) m, the wire forms an angle \( \theta \) with the vertical. The length of the wire \( L \) is 4.0 m.

To find \( \theta \), we can use the Pythagorean theorem. The vertical height of the chandelier remains approximately \( L \), while the horizontal displacement is \( x \). We can calculate the angle \( \theta \) using:

\[ \sin(\theta) = \frac{x}{L} \]

So,

\[ \theta = \sin^{-1}\left(\frac{0.17}{4.0}\right) \]

Calculating this,

\[ \theta \approx \sin^{-1}(0.0425) \approx 2.43^\circ \]

The horizontal force \( F_h \) is balanced by the horizontal component of the tension in the wire. The vertical force is balanced by the vertical component of the tension. Therefore,

\[ F_h = T \sin(\theta) \]

where \( T \) is the tension in the wire. To find \( T \), we use the fact that the vertical component of the tension equals the weight of the chandelier:

\[ T \cos(\theta) = mg \]

where \( m = 31 \) kg and \( g = 9.8 \, \text{m/s}^2 \). Thus,

\[ T = \frac{mg}{\cos(\theta)} \]

**2. Tension in the Wire:**

Let's calculate \( \cos(\theta) \):

\[ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} \]

\[ \cos(\theta) = \sqrt{1 - (0.0425)^2} \approx 0.9991 \]

So,

\[ T = \frac{31 \times 9.8}{0.9991} \approx 303.6 \, \text{N} \]

**3. Horizontal Force:**

Using the tension value, the horizontal force is:

\[ F_h = T \sin(\theta) \]

\[ F_h = 303.6 \times 0.0425 \approx 12.9 \, \text{N} \]

**Summary:**

- The horizontal force necessary to displace the chandelier 0.17 m to one side is approximately \( 12.9 \, \text{N} \).

- The tension in the wire is approximately \( 303.6 \, \text{N} \).