The downward force from the mass of the locomotive is the mass, m, 

 times gravitational acceleration, g, or 



F = mg = 104,500 kilograms (9.81 meters per second squared) 

 F = 1.025 x 10 6 newtons (2.3 x 10 5 pounds) 



Taking overturning moments about a rail, the center of mass of the 

 locomotive is equidistant from the two rails, or 0.76 meter (2.5 feet) 

 from the rail. The buoyancy and drag forces produce overturning moments 

 (+) and the mass of the locomotive a restraining force (-) . Summing 

 moments 



M = F D (0.76) + F n Z - F(0.76) 



D U 



M = 3.53 x 10 5 (0.76) + 5.24 x 10 5 (1.37) - 1.025 x 10 6 (0.76) 

 M = 2.07 x 10 5 newton-meters (1.48 x 10 5 foot-pounds) 



indicating that the overturning moments are greater than the restraining 

 moment. Therefore, the locomotive will be overturned. 



************************************* 

 ************** EXAMPLE PROBLEM 26 ************* 



GIVEN: A platform, 3 meters above ground level, is supported by square 

 columns with 14- by 14-centimeter (5.5 by 5.5 inches) cross sections. 

 A tsunami creates a surge with a depth of 2.44 meters (8 feet) under the 

 platform. The surge acts normal to the sides of the columns, which are 

 rigidly fixed at ground level. 



FIND : The moment of the surge force about the base of a column. 



SOLUTION : To determine the coefficient of drag, the columns may be con- 

 sidered as infinitely long columns, and from Table 6, C^ = 2.0. From 

 equation (318) 



u = 2./gh = 2/9.81(2.44) = 9.79 meters (32.1 feet) per second 

 The drag force on a column is given by equation (336) as 



F fl =pC A^= 1,026(2) (2.44) (0.14) IP^ZEL. 

 F = 3.36 x 10 4 newtons 



182 



