Calculating Horizontal Component of Tensile Force

This problem involves resolving a force vector into its components. A force acting at an angle can be broken down into a horizontal component and a vertical component. We are asked to find the horizontal component of the tensile force in the wire.

The given information is:

• Total tensile force, $F = 20 \text{ kN}$
• Angle the wire makes with the horizontal, $\theta = 60^\circ$

To find the horizontal component of a force that makes an angle $\theta$ with the horizontal, we use the formula:

Horizontal Component ($F_x$) $= F \times \cos(\theta)$

Here, $F = 20 \text{ kN}$ and $\theta = 60^\circ$.

We know that the value of $\cos(60^\circ)$ is $\frac{1}{2}$ or $0.5$.

Now, let's substitute the values into the formula:

$F_x = 20 \text{ kN} \times \cos(60^\circ)$

$F_x = 20 \text{ kN} \times \frac{1}{2}$

$F_x = 10 \text{ kN}$

Therefore, the horizontal component of the tensile force in the wire is $10 \text{ kN}$.

Let's compare this result with the given options:

Our calculated value of $10 \text{ kN}$ matches Option 1.

Revision Table: Key Concepts for Force Resolution

Additional Information on Force Components

Understanding how to resolve forces into components is fundamental in physics and engineering. It simplifies the analysis of forces acting at angles. For example, if you have multiple forces acting on an object at different angles, you can resolve each force into its horizontal and vertical components. Then, you can sum all the horizontal components to find the net horizontal force and sum all the vertical components to find the net vertical force.

The choice of using cosine or sine depends on the angle given. If the angle is given with respect to the horizontal axis, cosine gives the horizontal component and sine gives the vertical component. If the angle is given with respect to the vertical axis, then sine gives the horizontal component and cosine gives the vertical component.

Common angles and their cosine values often encountered in these types of problems include:

• $\cos(0^\circ) = 1$
• $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$
• $\cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707$
• $\cos(60^\circ) = \frac{1}{2} = 0.5$
• $\cos(90^\circ) = 0$

In this problem, the angle was $60^\circ$ with the horizontal, so using $\cos(60^\circ)$ was the correct approach for finding the horizontal component.