Question

What is the weight of the piece of aluminum shown below at 0.0974 lbs/in.^3 (hint:subtract the volume of the cylindrical hole.) Round to the nearest tenth as needed.

Answer 1

We will have the following:

First, we will determine the volume of the side plate:

`V_1=0.5\cdot4\cdot5\Rightarrow V_1=10`

Now, we find the volume of the front plate:

`V_2=0.5\cdot4\cdot4-\pi(2)^2(0.5)\Rightarrow V_2=8-2\pi`

Now, we calculate the total volume:

`V_t=2V_1+V_2\Rightarrow V_t=2(10)+(8-2\pi)`
`\Rightarrow V_t=28-2\pi`

Now, we calculate its weight:

`w=0.0974\cdot(b)/(in^3)\cdot(28-2\pi)\cdot in^3\Rightarrow w=2.115217751\ldots`
`\Rightarrow w\approx2.1`

So, it's weight is approximately 2.1 Lb.