Consider the diagram below. The box does not experience any motion in the vertical direction. An applied force FA acts on a box and an angle =40.0o. The box has a mass of 8.50 kg. What is the magnitude of the applied force if the box experiences a normal force of 125 N? Give your answer in N to 1 decimal place. Do NOT include units in your answer.

1 answer

To solve for the magnitude of the applied force \( F_A \) exerted at an angle of \( \theta = 40.0^\circ \), we can use the following steps:

  1. Identify the forces acting on the box.

    • The weight of the box \( W \) acts downwards: \[ W = m \cdot g = 8.50 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 83.435 , \text{N} \]
  2. The normal force \( N \) acts upwards and is given as 125 N.

  3. The applied force \( F_A \) can be split into its horizontal and vertical components. The vertical component \( F_{A_y} \) is given by: \[ F_{A_y} = F_A \cdot \sin(\theta) \] The horizontal component \( F_{A_x} \) is: \[ F_{A_x} = F_A \cdot \cos(\theta) \]

  4. Write the equation for the vertical forces. The box does not move vertically, so the upward forces equal the downward forces: \[ N + F_{A_y} = W \] Substituting \( F_{A_y} \): \[ N + F_A \cdot \sin(40.0^\circ) = W \]

  5. Rearranging for \( F_A \): \[ F_A \cdot \sin(40.0^\circ) = W - N \]

  6. Substitute the known values: \[ F_A \cdot \sin(40.0^\circ) = 83.435 , \text{N} - 125 , \text{N} \] \[ F_A \cdot \sin(40.0^\circ) = -41.565 , \text{N} \]

  7. Because the left side must equal the right side, we take the negative value indicating that the applied force is counteracting more than it provides in the vertical direction. Thus: \[ F_A \cdot \sin(40.0^\circ) = 41.565 , \text{N} \]

  8. Now solve for \( F_A \): \[ F_A = \frac{41.565 , \text{N}}{\sin(40.0^\circ)} \]

  9. Calculate \( \sin(40.0^\circ) \): \[ \sin(40.0^\circ) \approx 0.6428 \] Thus: \[ F_A = \frac{41.565}{0.6428} \approx 64.7 , \text{N} \]

Final answer: \[ \boxed{64.7} \]