Question

The density of steel is 490lb per feet^3. What is the weight of a piece of steel having the shape of an equilateral triangle with sides 3 feet long? The plate is 5/8 inches thick and has 2 holes with a 3 inch diameter drilled through it. I have no idea how to do this so a step by step on how to solve it would be great. The teacher said it could be plugged into excel to be solved but I have no idea how to do that either. THANKS!!

Answer 1

first find the volume of the triangle. since an equilateral triangle is 2 right triangles we can use (1/2) b*h, then times that by 2, since there are two right triangles to make one equilateral triangle.
b will be 3/2 = 1.5 and
h will be a^2 +1.5^2 = 3^2 (pathagorains therom. a^2+b^2=c^2) in our case h will equal a.

now the area is found we can find volume: v=area*depth, where depth is 5/8", but we have to convert inches to feet since our area is in units of feet.

now volume is found, but we now need to subtract the volume of the 2 drilled out holes
so
final volume = volume of triangle - (2)volume of drilled out holes
v = 0.134 cu ft - (2) 0.00256 = 0.1288 cu ft

now we do 0.1288 cu ft * 490 lb/cu ft = 63.15 lb