Question

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m 3 kg/m3 , length 81.2 cmcm and diameter 2.60 cmcm from a storage room to a machinist. Calculate the weight of the rod, www. Assume the free-fall acceleration is ggg = 9.80 m/s2m/s2 .

Answer 1

The weight of the rod is 32.87 N

Step-by-step explanation:

Density of the rod = 7800 kg/m

length of the rod = 81.2 cm = 0.812 m

diameter of rod = 2.60 cm = 0.026 m

acceleration due to gravity = 9.80 m/s^2

The rod can be assumed to be a cylinder.

The volume of the rod can be calculated as that of a cylinder, and can be gotten as

V =
`(\pi d^(2) l)/(4)`

where d is the diameter of the rod

l is the length of the rod

V =
`(3.142* 0.026^(2)* 0.812)/(4)` = 4.3 x 10^-4 m^3

We know that the mass of a substance is the density times the volume i.e

mass m = ρV

where ρ is the density of the rod

V is the volume of the rod

m = 4.3 x 10^-4 x 7800 = 3.354 kg

The weight of a substance is the mass times the acceleration due to gravity

W = mg

where g is the acceleration due to gravity g = 9.80 m/s^2

The weight of the rod W = 3.354 x 9.80 = 32.87 N