On this problem. We are looking at the torques applied by various forces on a 4 m long bar. Uh, that is attached to a pivot at one end. Ah, the force being applied in each case is 10 Newtons, and we're supposed to find the torque. In each case, we know that torque is equal to R F. Sign of fee. Uh, so all we didn't do is plug in our numbers and we should be able to figure out pretty quick here, Uh, our is in. It's at the end of ours. This before f is 10 and sign of fee. Uh, sign of 90 is, of course, one. So one and that will be equal to 40 Newton meters next. This one, it looks more like a J T. Uh, this is 120 degrees. Once again, it's at the end of the bar, 4 m away from the pivot. So that will be four times 10 times the sign of 1 20 which is going to be a 0.0 point 866 Yeah, and that will come out to 34.6 Newton meters. Uh, next once again, at the end of the bar. Uh, so that's 4 m. 10 Newtons. Same for each one times thesis. Ein of 30 is one half. So 10.5 and that will give us a total of 20 Newton meters. Uh Ah. This one is a little different. It is 60 degrees, but it's only halfway down the bars or 2 m away from the pivot. So the turkey will be equal to only two for the distance times the same 10 Newtons times the sign of 60 which is once again 0.86 six, which will get us a total of 17.3 Newton meters. Squeeze that in there. Okay, this one is directly on the pivot. So the eso the distance will be zero times 10 times are sign of 60 again is 0.866 But none of this matters because it's all being multiplied by zero. So this will be zero and then finally this one it is being, uh, directed at the end of the bar. So that will be again four times 10 and this. But the sign of 1 80 is zero. So once again, none of this matters. It will just be zero. So if you apply a force at the center of the pivot point or going toward the pivot point, no torque will be created.