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

When you push a wheelbarrow inclined at an angle to the horizontal, the applied force can be divided into two components: a **horizontal component** and a **vertical component**. To find the horizontal component of the force, you need to use the concept of resolving vectors.

The force of 150N is acting at an angle of 60º to the horizontal. The horizontal component of this force can be calculated using the cosine of the angle. The formula to determine the horizontal component \( F_{\text{horizontal}} \) is given by:

F_horizontal = F_applied \times \cos(\theta)

Where:

• F_applied is the magnitude of the applied force, which is 150N.
• \(\theta\) is the angle of inclination to the horizontal, which in this case is 60º.

Substitute the values into the formula:

F_horizontal = 150N \times \cos(60º)

We know that \(\cos(60º)\) equals 0.5.

Therefore:

F_horizontal = 150N \times 0.5 = 75N

Thus, the **horizontal component** of the applied force is 75N.