Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the force in each rod if the temperature is increased by 30° C. E copper = 120GPa, αcopper = 16.9x10-6/ ° C.

A)3.8196 * Newtons        B) 2.153 * Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as  120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = π.

Area Expansivity β =    

Area Expansivity β = 2α

α =  16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * =  0.0314

Area of Rod BC = π * = 0.0177.

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9* * 0.0314 * 30 = 3.183 * .

For Rod BC we have = 2 * 16.9* * 0.0177 * 30 = 1.794∈-5.

The forces on each rods will be given by.

Force AB = Pressure * Area  of Rod AB

                 = 120GPa * 3.183 * gives 3.8196 * Newtons.

Force BC = Pressure * Area of Rod BC

                = 120GPa * 1.794 * gives 2.153 * Newtons

no. dot product   =0 when vecs at right angles

volts times amps 6 x 0.5 3w