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?

1 answer

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).