Question

The accompanying diagram shows a 10.0 kilogram mass held at rest on a frictionless 30.0° incline by force F. Block mass=10.0 kg
What is the approximate magnitude of force F?

Answer 1

To find the magnitude of the force \( F \) holding the 10.0 kg mass at rest on a frictionless incline of 30.0°, we need to analyze the forces acting on the mass.

1. **Identify Forces**: The forces acting on the mass on the incline are:
- Gravitational force (\( W \)), which acts vertically down and can be calculated as:
\[
W = mg
\]
Where:
- \( m = 10.0 \, \text{kg} \) (mass)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

- The component of the gravitational force acting down the incline can be calculated using the sine of the incline angle:
\[
W_{\parallel} = W \sin(\theta)
\]
- The component of the gravitational force acting perpendicular to the incline (normal force) can be calculated using the cosine of the incline angle:
\[
W_{\perpendicular} = W \cos(\theta)
\]

2. **Calculate the Weight**:
\[
W = mg = 10.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 98.1 \, \text{N}
\]

3. **Calculate the Parallel Component of Weight**:
\[
W_{\parallel} = W \sin(30°) = 98.1 \, \text{N} \times 0.5 = 49.05 \, \text{N}
\]

4. **Determine the Force \( F \)**: Since the mass is at rest, the force \( F \) must balance the component of the weight acting down the incline:
\[
F = W_{\parallel} = 49.05 \, \text{N}
\]

Thus, the approximate magnitude of force \( F \) is **49.1 N** (rounded to three significant figures).