Mar 18th 2017, 07:54 AM
Join Date: Jun 2010
Location: Naperville, IL USA
For each of the three applied forces you need to determine the x- and y-components of those forces, then multiply each by the x- or y-distance, respectively,from point A, then sum up the results. To get you started: the 15 kN force is purely vertical, and its x-distance from A is 4m, so it contributes -15x4 = -60 kN-m of torque. It has a negative sign because it acts in a clockwise direction relative to A. Now consider the 21 kN force: it has components in both the x- and y-direction, so first you need to compute them. Then you need to determine the x- and y- offsets from point A. The x-distance is 4m+4m+3mcos(40), and the y-distance is -4m tan(15) +3m sin(40). Do you see why that is? OK, now you take it from here, and be careful as to whether each component of torque is positive (counter clockwise relative to point A) or negative (clockwise). Post back with your results and we'll review it for you.
Last edited by ChipB; Mar 19th 2017 at 02:11 PM.