Problem 2B-05**: The center of mass of a rectangular board

(a) Show that the center of mass of a uniform rectangular board is its center point.

(b) A uniform rectangular board, denoted as ABCD, is hinged at the midpoint of side AD. The midpoint is denoted as point E as shown in Fig. 1. The board can rotate freely around point E without any friction within the vertical plane of Fig. 1. A vertical string is attached to point C and a vertical upward force F applied to the string to hold the board such that the side CD forms an angle θ with a horizontal line. What is the magnitude of the force F?

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(a) The center point of the rectangular board is denoted as O, and through O a center line is drawn so that the rectangular board is seperated into two identical pieces as shown in Fig. 2. The righ-hand portion is then cut into N identical rectangular strips with a mass Δm for each strip. The strips are labeled as 1, 2, · · ·, N, starting from the one nearest to the center point O and move toward the right. By folding the right-hand portion of the rectangular board along the center line, all the strips in the right-hand portion are mapped into the left-hand portion. The strips in the left- hand protion are labeled as -1, -2, · · ·, -N. The k-th piece and the (-k)-th piece are mirror images. We denote the center of mass of k-th strip as Ck, and the center of mass of (-k)-th strip as C-k. As discussed in the previous problem 2B-04, the center of mass of a uniform rectangular strip is the center point of the strip. Thus point Ck, O and C-k lie on one line. Following the argument immediately after Eq. (5) in Topic 2B-02, the k-th strip can be reduced to a point particle of mass Δm located at Ck. Similar reduction can be made to the (-k)-th strip, too. Let the position vector from O to Ck be denoted as rk, and the position vector from O to C-k as r-k. From the symmetry it is easy to see that

                rk + r-k = 0 .

This means that

                (r1 + r-1) + (r2 + r-2) + · · ·, + (rN + r-N) = 0 .                                 (1)

Since each strip has an identical mass, Eq. (1) means that Eq. (4) of Topic 2B-02 is satisfied. Thus point O is the center of mass of the 2N-strip system, and thus O is the center of mass of the uniform rectangular board.

(b) From Eq. (6) of Topic 2B-02 we know that the torque τ generated by the gravitational pulls on the board around point E can be calculated as if a point-like particle of mass m is located at the center of mass point O. The situation is depicted in Fig. 2. The magnitude of torque τ is thus

                |τ| = b·mg·cosθ .                                 (2)

To calculate the torque τF that the force F generates around point E, we need to know the angle CEQ. The mid point of the side BC is denoted as point G. The angle GEC is denoted as φ. The angle GEQ is equal to θ. Thus

                ∠CEQ = ∠GEQ - ∠GEC = θ - φ.

Also we need to note that the length CE is √(a2 + 4b2) . Thus

                |τF| = |F|·{√(a2 + 4b2)}·cos(θ - φ) .

cosφ and sinφ can be expressed in terms of a and b by considering the triangle EGC. We get

                cosφ = EG / CE = 2b / {√(a2 + 4b2)},

                sinφ = CG / CE = a / √{(a2 + 4b2)},

so that

                cos(θ - φ) = cosθ·cosφ + sinθ·sinφ
                                = (2bcos θ + asinθ) / {√(a2 + 4b2)}.


                |τF| = |F|· (2bcos θ + asinθ).                                 (3)

Torque τ originates from point E and points perpendicularly into the paper, and torque τF originates from point E and points perpendicularly out of the paper. To make the board in Fig. 3 stationary, we must equate the right-hand sides of Eq. (2) and Eq. (3). Thus

                |F| = b·mg·cosθ .