WORLD FOR CIVIL

Statics & Strength of Materials

Example 3 - Torque Review Problems

Determine the magnitude, direction, and sense of the resultant forces shown. Determine where the resultant intersects the bottom of the body with respect to point O.

For the first part of this problem we sum the forces in both the x and y direction to determine the resultant components our final force vector will have.

Sum Fx = +500 lb. - 100 lb. = 400 lb. =Rx
Sum Fy = -400 lb. + 250 lb. = -150 lb. =Ry

We next find the resultant vector from it's components: R = square root (Rx²+Ry²) = Sqrt (400² + (-150)²)
R = 427 lb., and then direction from Tan f = Ry/Rx = -150/400 = -.375 The negative y-component and the positive x-component tell us that the resultant must be in the fourth quadrant. Solving for f we find f = -20.6 or 339.4 degrees.

To determine where the resultant intersects the bottom of the body with respect to point O, we realize that our resultant vector must produce the same torque with respect to point O as the orginal forces produce. Thus we must determine the resultant torque of our original forces.
Torque = + 100 lb. x 4" -400 lb. x 5" + 250 lb. x 17" - 500 lb. * 12" = -3350 in-lb. (clockwise direction)

Our resultant must produce the same torque when it is applied or acts at the bottom of the body. This actually simplifies the solution. At the bottom, the x-component of our vector (400 lb.) produces no torque as its line of action passes through point O. Only the y-component (-150 lb.,downward) produces a torque, and so we may write:
-150 lb. x (d") = -3350 in-lb., and therefore d = 22.3"