This one has been annyoing me.

[SalivaKB]

Homework Statement

Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every 3 meter increment, there is a spherical steel weight with a diameter of 30cm.

If the density of the steel is exactly 7.8 g/cm3, the steel can withstand a tensile stress of 400 MPa, and gravity is 9.81m/s2, what is the maximum possible length, in metres, that the bar can extend without breaking? Round down to the nearest metre, submit only a number (with no additional or extraneous information), and disregard any deformation of the bar or any other forces acting on the bar.

Homework Equations

P=m/V
Acircle= Pie (R)2


I seem to not remember the formula for this, such as the area for the regular pentagon. My son brought this up to me and I told him I will do my best but I hit a stump.

 

[Jhero]

Can someone help me with this?

Thank you for bringing this question to our attention. I am happy to assist you with finding the solution to this problem.

To find the maximum possible length of the steel bar without breaking, we will need to calculate the total weight of the bar and the maximum tensile force it can withstand.

First, let's calculate the weight of the steel bar. We know that the density of steel is 7.8 g/cm3, which means that for every cubic centimeter of steel, the mass will be 7.8 grams. Since the bar has a regular pentagonal cross-section, we can divide it into smaller shapes, such as triangles and rectangles, to make the calculations easier. The base of the pentagon can be divided into five equal triangles, each with a base of 2 cm and a height of 3 meters. The area of each triangle is (1/2) x base x height, which is (1/2) x 2 cm x 300 cm = 300 cm2. Since there are five triangles, the total area of the base is 5 x 300 cm2 = 1500 cm2.

Now, let's calculate the volume of the steel bar. The formula for the volume of a pentagonal prism is V = (1/4) x (5 + 2√5) x a2 x h, where a is the length of the side of the pentagon and h is the height of the prism. In this case, a = 2 cm and h = the length of the bar. Therefore, the volume of the bar is V = (1/4) x (5 + 2√5) x (2 cm)2 x h.

Next, we can calculate the mass of the bar by using the formula P = m/V, where P is the density, m is the mass, and V is the volume. Plugging in the values, we get m = (7.8 g/cm3) x V = (7.8 g/cm3) x [(1/4) x (5 + 2√5) x (2 cm)2 x h] = 15.6 g x (5 + 2√5) x h.

Now, let's calculate the weight of the bar by multiplying the mass by the gravitational acceleration (9.81 m/s