Question

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.

Answer 1

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: $$\boxed{{\displaystyle 64.7}}$$