An 3730 kg airplane has a center of mass located 32 m from the front of the plane. A 123 kg man sits 25 m from the front of the plane. Where should a 76 kg woman sit so that the center of mass of the loaded plane is the same as the empty airplane? (m from front)

Take moments about the centre of gravity (CG) the plane, so that the CG of the two persons is zero about the C from the front.

F........M...7...CG....X....W.......B

Take moments about the CG:
123 kg *7 m= 76 kg * X m
X = (123/76) * 7
= 11.33m (from the CG)

Distance from the front
=32+11.33 m
= 43.33 m from the front

Thanks a lot

To calculate the position where the 76 kg woman should sit in order to maintain the same center of mass as the empty airplane, we need to use the concept of torques. The torque is a measure of the turning effect of a force.

First, let's find the initial center of mass (CoM) of the empty airplane. Since the CoM is given as 32 m from the front of the plane, this is our initial reference point.

Next, we need to determine the new CoM with the man on board. We can do this by considering the torques produced by the airplane and the man.

The torque produced by an object is calculated by multiplying the force applied by the distance from the reference point. In this case, the force is the weight (mass * acceleration due to gravity) of the object.

The torque produced by the airplane is given by: T_airplane = (3730 kg) * (9.8 m/s^2) * (32 m).

The torque produced by the man is given by: T_man = (123 kg) * (9.8 m/s^2) * (25 m).

Since the total torque of the system should be zero to maintain rotational equilibrium, we can set up the equation:

T_airplane + T_man = 0.

(3730 kg) * (9.8 m/s^2) * (32 m) + (123 kg) * (9.8 m/s^2) * (25 m) = 0.

Now, let's solve this equation to find the new CoM position.

(3730 kg * 9.8 m/s^2 * 32 m) + (123 kg * 9.8 m/s^2 * 25 m) = 0.

(3730 * 9.8 * 32) + (123 * 9.8 * 25) = 0.

Using a calculator, we find that the sum of the torques is equal to -3,556,704 Nm.

Now, let's determine the distance the woman needs to sit from the front of the plane to neutralize the torque caused by the man. We'll call this distance x.

The torque produced by the woman is given by: T_woman = (76 kg) * (9.8 m/s^2) * (x m).

The total torque of the system, including the woman, should now be zero, so we set up the equation:

T_airplane + T_man + T_woman = 0.

(3730 kg) * (9.8 m/s^2) * (32 m) + (123 kg) * (9.8 m/s^2) * (25 m) + (76 kg) * (9.8 m/s^2) * (x m) = 0.

Substituting the values we have, the equation becomes:

(3730 * 9.8 * 32) + (123 * 9.8 * 25) + (76 * 9.8 * x) = 0.

Simplifying further, we have:

-3556704 + 301940 - 752.8x = 0.

Combine like terms:

-3,556,704 + 301,940 - 752.8x = 0.

-3,254,764 - 752.8x = 0.

To solve for x, divide both sides by -752.8:

x = -3,254,764 / -752.8.

Using a calculator, we find that x ≈ 4,328.1 m.

Thus, the woman should sit approximately 4,328.1 m from the front of the plane in order to maintain the same center of mass as the empty airplane.