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Give step-by-step solution with explanation and final answer:15. A man pries up one end of a 480-Ib crate with a steel pipe 6 ft long. If he exerts a force of 80 Ib, the distance of the fulcrum from the crate is a. 6in. b. 9 in. c. 18in. d. 24 in.

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Give step-by-step solution with explanation and final answer:Uploaded Image15. A man pries up one end of a 480-Ib crate with a steel pipe 6 ft long. If he exerts a force of 80 Ib, the distance of the fulcrum from the crate is a. 6in. b. 9 in. c. 18in. d. 24 in.

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Asked by: Test User 100252
Created at: 2025-05-21 18:48:28

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Given: - Weight of crate (load, \( F_L \)) = **480 lb** - Length of steel pipe = **6 ft = 72 in** (since \(1 \text{ ft} = 12 \text{ in}\)) - Effort applied (\( F_E \)) = **80 lb** - Distance from fulcrum to crate = \( x \) (what we need to find) - Distance from fulcrum to where effort is applied = \( 72 - x \) (since total length is 72 in) We use the **principle of levers** (Law of the Lever): \[ \text{Effort} \times \text{Effort Arm} = \text{Load} \times \text{Load Arm} \] \[ 80 \times (72 - x) = 480 \times x \] Let's solve for \( x \): 1. **Expand the equation:** \[ 80 \times 72 - 80x = 480x \] \[ 5760 - 80x = 480x \] 2. **Combine like terms:** \[ 5760 = 480x + 80x \] \[ 5760 = 560x \] 3. **Solve for \( x \):** \[ x = \frac{5760}{560} \] \[ x = 10.2857 \text{ in} \] None of the choices exactly match, but let's check which is **closest**: Choices: - a. 6 in - b. 9 in - c. 18 in - d. 24 in **9 in** is the closest answer to 10.3 in. --- ### **Final Answer** **b. 9 in** --- #### **Explanation:** We balanced the moments (force × distance) around the fulcrum, solved for the unknown distance from the fulcrum to the crate, and matched it to the closest given answer.

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