Question

Two people are carrying a uniform wooden board that is 3.40 m long and weighs 160 N.

If one person applies an upward force equal to 60.0 N at one end, at what point does the other person lift?
d=? Units= m from the end where the first person lifts

Please help !!! I have tried everything and keep getting the wrong answer and only have 1 attempt left !

Answer 1

First of all, the force of the other person should be F2 = W - F1 = 160N - 60N = 100N.
Let d1 and d2 be the distance between the persons to the center of gravity of the board. Since the magnitude of the torque by the two are the same, we haveF1*d1 = F2*d2. So d2 = F1*d1/F2 = 60N*(3.4m/2)/100N = 1.02m. So d = d1+d2 = 1.7m+1.02m = 2.72m.

Answer 2

We know that the other person applies an upward force of 100N, because the forces must add up to 160N. We also know that the center of gravity ofthe board is in the middle, at 1.70m from each end.

The torque from first person is 60N*1.70m=102Nm.

So, the torque from the second person must be 102Nm, too.

Let that person lift at distance d from the center. Then, his torque is 100N*d = 102Nm.

we divide to find d=1.02m. The force is applied 1.02 meters from the center, so it's applied 1.70-1.02 = 0.68m from the end of the plank.