A wheelbarrow inclined at 60º to the horizontal is pushed with a force of 150N. What is the horizontal component of the applied force

1. Analyze the question

We are given a force ($F$) of 150N applied at an angle ($\theta$) of 60º to the horizontal. We need to find the horizontal component of this force.

2. Explain the concept

In physics, when a force is applied at an angle, it can be resolved into two perpendicular components:

• Horizontal component ($F_x$): This is the part of the force acting along the x-axis (the ground). It is calculated using the cosine function: $$F_x = F \cos(\theta)$$
• Vertical component ($F_y$): This is the part of the force acting along the y-axis (upwards). It is calculated using the sine function: $$F_y = F \sin(\theta)$$

Since the question asks for the horizontal component, we will use the cosine formula.

3. Show the solution

• Given: • Force ($F$) = 150N • Angle ($\theta$) = 60º
• Formula: • $F_x = F \cos(60º)$
• Calculation: • We know that $\cos(60º) = 0.5$ (or $\frac{1}{2}$) • $F_x = 150 \times 0.5$ • $F_x = 75\text{N}$

4. Identify the answer

Comparing our result to the given options: A. 130N B. 75N C. 60N D. 50N

The calculated value matches option B.

5. Tip

A simple way to remember which trig function to use: "Cos is close." The horizontal component is usually the one "closer" to the horizontal axis (the adjacent side of the triangle), so it always pairs with cosine. If you are ever confused, just remember CAH (Cosine = Adjacent / Hypotenuse).