Question

1. using material densities from the following table. what is the total weight (in pounds) of a 6-foot-tall by 15-foot-long wall that is constructed of 4-inch lightweight concrete masonry units with a layer of ½ plywood on each side?

Explanation:

To solve this problem, we need the material densities from the table (which is not provided here). However, we can outline the general steps to calculate the total weight:

Step 1: Calculate the volume of each material

• Lightweight Concrete Masonry Units (CMU):

The wall is 6 feet tall, 15 feet long, and 4 inches thick. First, convert the thickness to feet: \( 4 \text{ inches} = \frac{4}{12} \text{ feet} = \frac{1}{3} \text{ feet} \).
Volume of CMU: \( V_{\text{CMU}} = \text{height} \times \text{length} \times \text{thickness} = 6 \times 15 \times \frac{1}{3} \).

• Plywood (both sides):

Each plywood layer is \( \frac{1}{2} \) inch thick. Convert thickness to feet: \( \frac{1}{2} \text{ inch} = \frac{1/2}{12} \text{ feet} = \frac{1}{24} \text{ feet} \). There are 2 layers (one on each side).
Volume of one plywood layer: \( V_{\text{plywood, single}} = 6 \times 15 \times \frac{1}{24} \).
Total volume of plywood: \( V_{\text{plywood}} = 2 \times V_{\text{plywood, single}} \).

Step 2: Multiply volume by density to get weight

• Let \(

ho_{\text{CMU}} \) = density of Lightweight CMU (in pounds per cubic foot, \( \text{lb/ft}^3 \))
Let \(
ho_{\text{plywood}} \) = density of plywood (in \( \text{lb/ft}^3 \))

Weight of CMU: \( W_{\text{CMU}} = V_{\text{CMU}} \times
ho_{\text{CMU}} \)
Weight of plywood: \( W_{\text{plywood}} = V_{\text{plywood}} \times
ho_{\text{plywood}} \)

Step 3: Sum the weights

Total weight: \( W_{\text{total}} = W_{\text{CMU}} + W_{\text{plywood}} \)

Example (assuming typical densities):

• Typical density of Lightweight CMU: \(

ho_{\text{CMU}} \approx 80 \, \text{lb/ft}^3 \)

• Typical density of plywood: \(

ho_{\text{plywood}} \approx 35 \, \text{lb/ft}^3 \)

Calculate \( V_{\text{CMU}} \):

\( V_{\text{CMU}} = 6 \times 15 \times \frac{1}{3} = 30 \, \text{ft}^3 \)

Calculate \( W_{\text{CMU}} \):

\( W_{\text{CMU}} = 30 \times 80 = 2400 \, \text{lb} \)

Calculate \( V_{\text{plywood, single}} \):

\( V_{\text{plywood, single}} = 6 \times 15 \times \frac{1}{24} = \frac{90}{24} = 3.75 \, \text{ft}^3 \)

Calculate \( V_{\text{plywood}} \):

\( V_{\text{plywood}} = 2 \times 3.75 = 7.5 \, \text{ft}^3 \)

Calculate \( W_{\text{plywood}} \):

\( W_{\text{plywood}} = 7.5 \times 35 = 262.5 \, \text{lb} \)

Total weight:

\( W_{\text{total}} = 2400 + 262.5 = 2662.5 \, \text{lb} \)

Final Answer (with typical densities):

\( \boldsymbol{2662.5} \) pounds (note: actual answer depends on the specific densities from the table).

Answer:

To solve this problem, we need the material densities from the table (which is not provided here). However, we can outline the general steps to calculate the total weight:

Step 1: Calculate the volume of each material

• Lightweight Concrete Masonry Units (CMU):

The wall is 6 feet tall, 15 feet long, and 4 inches thick. First, convert the thickness to feet: \( 4 \text{ inches} = \frac{4}{12} \text{ feet} = \frac{1}{3} \text{ feet} \).
Volume of CMU: \( V_{\text{CMU}} = \text{height} \times \text{length} \times \text{thickness} = 6 \times 15 \times \frac{1}{3} \).

• Plywood (both sides):

Each plywood layer is \( \frac{1}{2} \) inch thick. Convert thickness to feet: \( \frac{1}{2} \text{ inch} = \frac{1/2}{12} \text{ feet} = \frac{1}{24} \text{ feet} \). There are 2 layers (one on each side).
Volume of one plywood layer: \( V_{\text{plywood, single}} = 6 \times 15 \times \frac{1}{24} \).
Total volume of plywood: \( V_{\text{plywood}} = 2 \times V_{\text{plywood, single}} \).

Step 2: Multiply volume by density to get weight

• Let \(

ho_{\text{CMU}} \) = density of Lightweight CMU (in pounds per cubic foot, \( \text{lb/ft}^3 \))
Let \(
ho_{\text{plywood}} \) = density of plywood (in \( \text{lb/ft}^3 \))

Weight of CMU: \( W_{\text{CMU}} = V_{\text{CMU}} \times
ho_{\text{CMU}} \)
Weight of plywood: \( W_{\text{plywood}} = V_{\text{plywood}} \times
ho_{\text{plywood}} \)

Step 3: Sum the weights

Total weight: \( W_{\text{total}} = W_{\text{CMU}} + W_{\text{plywood}} \)

Example (assuming typical densities):

• Typical density of Lightweight CMU: \(

ho_{\text{CMU}} \approx 80 \, \text{lb/ft}^3 \)

• Typical density of plywood: \(

ho_{\text{plywood}} \approx 35 \, \text{lb/ft}^3 \)

Calculate \( V_{\text{CMU}} \):

\( V_{\text{CMU}} = 6 \times 15 \times \frac{1}{3} = 30 \, \text{ft}^3 \)

Calculate \( W_{\text{CMU}} \):

\( W_{\text{CMU}} = 30 \times 80 = 2400 \, \text{lb} \)

Calculate \( V_{\text{plywood, single}} \):

\( V_{\text{plywood, single}} = 6 \times 15 \times \frac{1}{24} = \frac{90}{24} = 3.75 \, \text{ft}^3 \)

Calculate \( V_{\text{plywood}} \):

\( V_{\text{plywood}} = 2 \times 3.75 = 7.5 \, \text{ft}^3 \)

Calculate \( W_{\text{plywood}} \):

\( W_{\text{plywood}} = 7.5 \times 35 = 262.5 \, \text{lb} \)

Total weight:

\( W_{\text{total}} = 2400 + 262.5 = 2662.5 \, \text{lb} \)

Final Answer (with typical densities):

\( \boldsymbol{2662.5} \) pounds (note: actual answer depends on the specific densities from the table).