WORLD FOR CIVIL

Additional Important Examples

Some additional examples with brief solutions

1. Three vectors are shown in the diagram below. Find the resultant vector (magnitude and direction):

R = -2 A - 2 B -3 C

2. A block has four forces acting on it as shown in the diagram.

Determine the resultant (net) torque (magnitude & direction) about the center of mass (point O).

Determine the resultant force (magnitude & direction) and where along the base it would act to produce the same torque about point C

Sum Torque-c: +64.3 lb * 3 ft – 76.6 lb * 3 ft -128.6 lb * 2 ft – 153 lb * 3 ft + 100 lb * 2ft – 200 lb * 2 ft = - 953 ft-lb
Sum Force x = -64.3 lb + 126.6 lb + 100 lb = 162.3 lb
Sum Force y = 76.6 lb – 153.2 lb + 200 lb = 123. 4 lb

R = sqrt( 162.3 ^2 + 123.4^2) = 204 lb. Tan (Theta) = 123.4/162.3 ; Theta = 37.2 deg

123.4 lb (x ft) – 162.3 lb * (3 ft) = -953 ft-lb, so x = - 3.78 ft (to left of center of bottom)

3. The beam shown below is supported by a roller at point B and by a pinned joint at point E. Determine the support forces at points B and E

Solution: Sum Fx: 900 lb + 960 lb – Ex = 0

Sum Fy: - (600 lb/ft * 5ft) + By – 720 lb – (1000 lb/ft * 5 ft) + Ey = 0

Sum Torque B: +(3000 lb * 2.5 ft) – (720 lb * 5 ft) – (5000 lb * 12.5 ft) + Ey * 15 ft = 0

Ex = 1860 lb (negative x direction); Ey = + 3907 lb; By = + 4813 lb