MN 219 Test 2 Chapter 3 Name Problem 1 Determine the magnitu

MN 219 Test 2 Chapter 3 Name: Problem 1: Determine the magnitude, direction, and sense of the resultant force of the force system shown below Locate the resultant with res 10 points t to point A. Neglect the weight and thickness of the beam. 40 k 10k 45 2.4 k/ft 2 k/ft N.T.S

Solution

Total vertical load on beam = 0.5*5*2.4 + 40+10*sin45 + 2*5 = 63.07 kips(vertically downwards)

Total horizontal load on beam = 10*cos45 = 7.07 kips(towards left)

magnitude of resultant force = sqrt(63.07²+7.07²)=63.46 kips

the angle resultant force makes with positive x axis = 180+tan^-1(63.07/7.07)=263.6^o anticlockwise from positive x axis

location of resultant force measured from A = [(0.5*2.4*5*2*5/3)+(40*10)+(10*sin45*13)+(2*5*18.5)]/63.07 = 11.05 ft