**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, **F**_{A},
**F**_{B} and
**F**_{C}, 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 **F**_{A}, **F**_{B} and
**F**_{C} are expressed by their respective x, y and z components as

**F**_{A} = (-1, -1, 0), **F**_{B} =
(2, -1, 0) and **F**_{C} = (-1, -2, 0)

respectively. Suppose a force **F**_{D} is applied at point D to make
the board stationary, what should **F**_{D} be?

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solution

**Solution:**

Some will rush to the temptation of the equation **F**_{A} +
**F**_{B} + **F**_{C} + **F**_{D} = 0 and think
that **F**_{D} can be calculated easily from this equation. However,
this approach is wrong since the sum {**F**_{A} +
**F**_{B} + **F**_{C} + **F**_{D}} does not
need to become zero. A non-zero sum of {**F**_{A} +
**F**_{B} + **F**_{C} + **F**_{D}} 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 {**F**_{A} +
**F**_{B} + **F**_{C} + **F**_{D}}. 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
**F**_{A}, **F**_{B}, **F**_{C} and
**F**_{D} 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 **r**_{A},
**r**_{B}, **r**_{C} and **r**_{D} respectively
as shown in Fig. 2. We have

**τ** = **τ**_{A} + **τ**_{B} +
**τ**_{C} + **τ**_{D},

where

**τ**_{A} = **r**_{A} x **F**_{A},

**τ**_{B} = **r**_{B} x **F**_{B},

**τ**_{C} = **r**_{C} x **F**_{C},

and

**τ**_{D} = **r**_{D} x **F**_{D}

respectively. The force **F**_{D} can be decomposed into two
mutually perpendicular components both of that lie in the x-y plane;
**F**_{D}^{⊥} is perpendicular to
**r**_{D}, and **F**_{D}^{//} is parallel to
**r**_{D}. Using the property of vector products
that **r**_{D} x **F**_{D}^{//} = 0, we have

**τ** = **r**_{A} x **F**_{A} +
**r**_{B} x **F**_{B} +
**r**_{C} x **F**_{C} +
**r**_{D} x **F**_{D}^{⊥} .
(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 **r**_{A},
**r**_{B}, **r**_{C} and **r**_{D}
respectively. Thus we get

**r**_{A} x **F**_{A} =
{0, 0, (-3)·(-1) - 1·(-1)} = (0, 0, 4) ,
(2)

**r**_{B} x **F**_{B} =
{0, 0, (-1)·(-1) - (-2)·2} = (0, 0, 5),
(3)

and

**r**_{C} x **F**_{C} =
{0, 0, 3·(-2) - (-1)·(-1)} = (0, 0, -7) .
(4)

Let us write **F**_{D}^{⊥} = (*a*, *b*, 0).
Then

**r**_{D} x **F**_{D}^{⊥} =
(0, 0, 2*b* - 4*a*) .

Since **r**_{D} ⊥ **F**_{D}^{⊥}, we must have
**r**_{D} · **F**_{D}^{⊥} = 0.
Since
**r**_{D} · **F**_{D}^{⊥}
= (2*a* + 4*b*), we get *a* = -2*b*. This leads to

**r**_{D} x **F**_{D}^{⊥} =
(0, 0, 10*b*) .
(5)

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

**τ** = (0, 0, 4 + 5 - 7 + 10*b*) .

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

**F**_{D}^{⊥} = (2/5, -1/5, 0).

It should be noted that **F**_{D}^{//} does not contribute to
this calculation so it can be anything. This means that **F**_{D} can
not be determined uniquely. Only thing we can say is what
**F**_{D}^{⊥} should be.