Question

practice weight and balance ques question 31 of 50
two boxes which weigh 10 pounds and 5 pounds are placed in an airplane so that
their distance off from the cg are 4 feet and 2 feet respectively. how far forward
of the cg should a third box, weighing 20 pounds, be placed so that the cg will not
be changed?
a 3.4 feet.
b 2.8 feet.
c 3 feet.

Explanation:

Step1: Set up the moment - balance equation

The principle of moments states that for the center of gravity (CG) not to change, the sum of the moments before and after adding the third box must be equal. The moment of a force (or weight in this case) is given by \(M = w\times d\), where \(w\) is the weight and \(d\) is the distance from the CG.
Let the distance of the third box from the CG be \(x\). The sum of the moments of the first two boxes is \(10\times4+5\times2\). The sum of the moments of all three boxes is \(10\times4 + 5\times2+20\times x\). Since the CG does not change, the moment of the third box alone must be zero (because the total moment before adding the third box is balanced by the total moment after adding the third box, and the contribution of the first two boxes remains the same in the balance condition). Mathematically, using the moment - balance formula \(w_1d_1+w_2d_2=w_1d_1+w_2d_2 + w_3d_3\) (for CG non - change), we can also think in terms of the fact that the moment of the added weight must be zero. Another way is to use the formula for the balance of moments: \(10\times4+5\times2= (10 + 5+20)\times d_{new}\) (but since the original moment \(M_0=10\times4 + 5\times2\) and we want to find \(x\) such that the new moment is the same as the original moment. In fact, we can use the formula \(10\times4+5\times2=10\times4+5\times2+20x\) (which simplifies to \(20x = 0\) in the balance sense, but a better approach is using the formula \(w_1d_1+w_2d_2=w_3d_3\) for the balance of the additional weight. Wait, actually, the correct formula is based on the fact that the moment of the weights about the CG should balance. Let's assume the original moment (without the third box) is \(M_1=10\times4+5\times2\). After adding the third box, for the CG not to change, the moment of the third box must cancel out the change. But a more straightforward formula is \(10\times4+5\times2=20x\).
Calculate \(10\times4+5\times2\):

$$
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$$

So the equation becomes \(50=20x\).

Step2: Solve for \(x\)

We have the equation \(20x=50\). Using the formula \(x=\frac{w_1d_1 + w_2d_2}{w_3}\) (where \(w_1 = 10\), \(d_1=4\), \(w_2 = 5\), \(d_2 = 2\), \(w_3=20\)).

$$x=\frac{10\times4+5\times2}{20}=\frac{40 + 10}{20}=\frac{50}{20}=2.5$$

Answer:

B. \(2.5\) feet.