Problem 2B-03**: Balancing a horizontal board

A board is placed on a horizontal surface as shown in Fig. 1. There is a hole at point O and the board is fitted at point O to a pole that is fixed on the horizontal surface. There is no resistance for the board to rotate around the pole at point O. Three forces, FA, FB and FC, are applied to points A, B, and C respectively. A rectangular coordinate system is chosen with point O as its origin and the z-axis coming out of the paper, as shown in Fig. 1. The rectangular coordinate representations of points A, B, C and D, in the order of x, y and z coordinate values, are

                A = (-3, 1, 0),     B = (-1, -2, 0),     C = (3, -1, 0),     and     D = (2, 4, 0)

respectively. The forces FA, FB and FC are expressed by their respective x, y and z components as

                FA = (-1, -1, 0),     FB = (2, -1, 0)     and     FC = (-1, -2, 0)

respectively. Suppose a force FD is applied at point D to make the board stationary, what should FD be?






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Solution:

Some will rush to the temptation of the equation FA + FB + FC + FD = 0 and think that FD can be calculated easily from this equation. However, this approach is wrong since the sum {FA + FB + FC + FD} does not need to become zero. A non-zero sum of {FA + FB + FC + FD} tries to move the board, including the hole at point O, away from its original configuration. The pole will prevent the hole to move by supplying a counter force necessary to cancel out the sum of {FA + FB + FC + FD}. Thus the consideration of the balance of the giving forces will not lead us anywhere.

What we need to consider is the torques generated by the forces FA, FB, FC and FD around point O respectively. Since there is no friction in rotating the board around the pole at point O, the sum of the above mentioned four torques should be zero in order to keep the board stationary. Let position vectors OA, OB, OC and OD be denoted as rA, rB, rC and rD respectively as shown in Fig. 2. We have

                τ = τA + τB + τC + τD,

where

                τA = rA x FA,

                τB = rB x FB,

                τC = rC x FC,

and

                τD = rD x FD

respectively. The force FD can be decomposed into two mutually perpendicular components both of that lie in the x-y plane; FD is perpendicular to rD, and FD// is parallel to rD. Using the property of vector products that rD x FD// = 0, we have

                τ = rA x FA + rB x FB + rC x FC + rD x FD .                                 (1)

Since point O is the origin, the given coordinate components of point A, B, C and D become directly the components of position vectors rA, rB, rC and rD respectively. Thus we get

                rA x FA = {0, 0, (-3)·(-1) - 1·(-1)} = (0, 0, 4) ,                                 (2)

                rB x FB = {0, 0, (-1)·(-1) - (-2)·2} = (0, 0, 5),                                 (3)

and

                rC x FC = {0, 0, 3·(-2) - (-1)·(-1)} = (0, 0, -7) .                                 (4)

Let us write FD = (a, b, 0). Then

                rD x FD = (0, 0, 2b - 4a) .

Since rDFD, we must have rD · FD = 0. Since rD · FD = (2a + 4b), we get a = -2b. This leads to

                rD x FD = (0, 0, 10b) .                                 (5)

Substituting Eqs. (2), (3), (4) and (5) into Eq. (1), we get

                τ = (0, 0, 4 + 5 - 7 + 10b) .

τ = 0 then leads to 2 + 10b = 0, or b = -1/5 and a = -2b = 2/5. Thus we have

                FD = (2/5, -1/5, 0).

It should be noted that FD// does not contribute to this calculation so it can be anything. This means that FD can not be determined uniquely. Only thing we can say is what FD should be.