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:
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Identify the forces acting on the box.
- The weight of the box \( W \) acts downwards: \[ W = m \cdot g = 8.50 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 83.435 , \text{N} \]
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The normal force \( N \) acts upwards and is given as 125 N.
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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: \[ F_{A_x} = F_A \cdot \cos(\theta) \]
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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 \( F_{A_y} \): \[ N + F_A \cdot \sin(40.0^\circ) = W \]
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Rearranging for \( F_A \): \[ F_A \cdot \sin(40.0^\circ) = W - N \]
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Substitute the known values: \[ F_A \cdot \sin(40.0^\circ) = 83.435 , \text{N} - 125 , \text{N} \] \[ F_A \cdot \sin(40.0^\circ) = -41.565 , \text{N} \]
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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: \[ F_A \cdot \sin(40.0^\circ) = 41.565 , \text{N} \]
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Now solve for \( F_A \): \[ F_A = \frac{41.565 , \text{N}}{\sin(40.0^\circ)} \]
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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: \[ \boxed{64.7} \]