1. **State the problem:** We have two boxes on an inclined ramp. The lower box has a mass of 48.0 kg, and the upper box has a mass $m = 32$ kg. The ramp has a vertical height of 2.50 m and a horizontal base length of 4.75 m. We want to analyze forces or related quantities on the ramp. 2. **Calculate the length of the ramp (hypotenuse):** Using the Pythagorean theorem: $$\text{length} = \sqrt{(2.50)^2 + (4.75)^2} = \sqrt{6.25 + 22.56} = \sqrt{28.81} = 5.37\,\text{m}$$ 3. **Calculate the angle of the incline $\theta$:** $$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2.50}{5.37} = 0.4657$$ $$\theta = \arcsin(0.4657) \approx 27.8^\circ$$ 4. **Explain the forces on the boxes:** - The weight of each box acts vertically downward. - The component of weight parallel to the ramp is $mg \sin \theta$. - The component of weight perpendicular to the ramp is $mg \cos \theta$. 5. **Calculate the parallel component of weight for each box:** - For the lower box (48.0 kg): $$F_{\text{parallel, lower}} = 48.0 \times 9.8 \times \sin(27.8^\circ) = 48.0 \times 9.8 \times 0.4657 = 219.3\,\text{N}$$ - For the upper box (32 kg): $$F_{\text{parallel, upper}} = 32 \times 9.8 \times 0.4657 = 146.1\,\text{N}$$ 6. **Calculate the total parallel force due to both boxes:** $$F_{\text{total parallel}} = 219.3 + 146.1 = 365.4\,\text{N}$$ 7. **Summary:** The ramp length is approximately 5.37 m, the incline angle is about 27.8 degrees, and the total force pulling the boxes down the ramp parallel to its surface is approximately 365.4 N. This analysis helps understand the forces acting on the boxes on the inclined plane.