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The problem states:

- A force of \(90 \, \text{N}\) is applied to a lawnmower along a handle that makes a \(30^\circ\) angle with the vertical.

**a.** What is the magnitude of the component of this force that propels the lawnmower?
**b.** What is the magnitude of the component of this force that keeps the lawnmower in contact with the ground?

Let's break this down:

### Part a: Component that propels the lawnmower

The force propelling the lawnmower is along the horizontal direction. Since the handle makes a \(30^\circ\) angle with the vertical, the angle with the horizontal is \(60^\circ\) (because \(90^\circ - 30^\circ = 60^\circ\)).

The horizontal component \(F_{\text{horizontal}}\) is given by:
\[
F_{\text{horizontal}} = F \cdot \cos(\theta)
\]
Where:
- \(F = 90 \, \text{N}\) is the total force.
- \(\theta = 60^\circ\) is the angle with the horizontal.

\[
F_{\text{horizontal}} = 90 \cdot \cos(60^\circ)
\]
\[
F_{\text{horizontal}} = 90 \cdot 0.5 = 45 \, \text{N}
\]

### Part b: Component that keeps the lawnmower in contact with the ground

The vertical component of the force is what pushes the lawnmower downwards, keeping it in contact with the ground. This component is calculated using the sine function because it is perpendicular to the ground:

\[
F_{\text{vertical}} = F \cdot \sin(\theta)
\]
Where \(\theta = 60^\circ\) (the angle with the horizontal):
\[
F_{\text{vertical}} = 90 \cdot \sin(60^\circ)
\]
\[
F_{\text{vertical}} = 90 \cdot \frac{\sqrt{3}}{2} \approx 90 \cdot 0.866 = 77.94 \, \text{N}
\]

### Final Answers:
- The component that propels the lawnmower is \(45 \, \text{N}\).
- The component that keeps the lawnmower in contact with the ground is approximately \(77.94 \, \text{N}\).

Let me know if you'd like further details or clarifications!

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Here are 5 related questions to explore this concept further:
1. How would the components change if the angle with the vertical were 45 degrees instead?
2. What is the total work done by the force if the lawnmower is pushed for 10 meters horizontally?
3. How would you calculate the net force if friction were introduced?
4. How does the angle between the handle and the vertical affect the efficiency of pushing the lawnmower?
5. What if the force applied were 120 N instead of 90 N? How would that affect the components?

**Tip:** Always consider breaking a force into components when dealing with angles—it simplifies the analysis!

A force of 90 N is applied to a lawnmower along a handle that makes a 30° angle with the vertical.
a. What is the magnitude of the component of this force that propels the lawnmower?
b. What is the magnitude of the component of this force that keeps the lawnmower in contact with the ground?

Solve a problem where a 90 N force is applied to a lawnmower along a handle at a 30° angle with the vertical. Learn how to calculate the horizontal and vertical components of the force using trigonometric functions. Suitable for students in Grades 9-11.

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