Verified answer

If you draw a diagram of the see-saw, and the angle is sloping up towards mass C, then the torque from mass A is -60kg*1m*g*cos(30º) and the torque from mass C is 10kg*3m*g*cos(30º). For equilibrium all torques must sum to zero, so add the torque from A to the torque from C (observing the sign difference). If that sum is not zero, the torque from B must equal the negative of the remainder. (Since g and cos30º appears in all terms, we can compute the balance without them.)

∆T = 10kg*3m -60kg*1m = -30

r*30 = 30, r = 1m; the weight is then on the same side as weight C.

For the second part, remember that torque is the force (mass times g for weights) times the distance along a line drawn perpendicular to the direction of the force and through the pivot.

Note regarding the angle: If the angle between the force vector and vector from axis to point where force is applied is ø, then the torque is F*r*sinø. However, in this problem, the angle is given as 30º from the *horizontal*, but the force (weight) is *vertical*, so the angle is 90º - ø, and the torque becomes F*r*cosø.

1) moment = Force X perpendicular distance from the pivot

moments of A and c .

moments of A = 60(9.8)(1)sin30=294 Nm

moments of C= 10(9.8)(3sin30) =147 Nm

294-147 = 147 Nm .

Now to balance be must provide moment of 147 Nm

30(9.8)xsin30 = 147 .

x=1 m from pivot .

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