At 20 c, a steel rod of length 40.000 cm and a brass rod
oflength 30.000 cm, both of the same diameter, are placed end to
endbetween two rigid supports 70.000 cm apart, with no initial
stressin the rods. the temperature of the rods is now raised to
60c. (a) what is the stress (the same in both rods) at 60 c. (b)
howfar from its old positioin is the new junction between
therods?

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

Step 1: identify the given parameters and standard parameters

⇒Steel rod: length of rod = 40.000 cm

                    Young modulus(Υ) = 20 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

                     stress in the rod =?

 ⇒Brass rod: length of rod = 30.000 cm

                    Young modulus(Υ) = 9 X 10¹⁰ pa

                    coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

                     stress in the rod =?

Step 2: calculate the stress in each rod

⇒Steel rod: stress in the rod = -Y*α*ΔT

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

                                                =  -9.6 X 10⁷ pa

⇒Brass rod: stress in the rod = -Y*α*ΔT

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

                                                = (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

                                                =  -7.2 X 10⁷ pa

∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

Step 3: calculate the new length of each rod

⇒Steel rod: New length = ΔL + L₀

                                   ΔL = α*L₀*ΔT

                                  ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

                                   ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

                    New length = ΔL + L₀ = 0.0192cm + 40cm

                    New length = 40.0192cm

⇒Brass rod: New length = ΔL + L₀

                                    ΔL = α*L₀*ΔT

                                   ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

                                   ΔL = 2400 X 10⁻⁵ cm = 0.024cm

                    New length = ΔL + L₀ = 0.024cm + 30cm

                    New length = 30.024cm

Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

f=ma newton

f=100x2.5=250n