Understanding the Moment of Force on a Spacecraft

The question asks for the magnitude of the moment (also known as torque) acting on a spacecraft. The moment of a force is a measure of its tendency to cause a body to rotate about a specific point or axis. It is calculated as the cross product of the position vector from the point of rotation to the point where the force is applied, and the force vector itself.

In this problem, the spacecraft is located at a position given by the vector $\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k}$. The force applied to the spacecraft is given by the vector $\vec{F} = \lambda \hat{k}$. The moment is calculated about the origin, which is the point (0, 0, 0). Therefore, the position vector $\vec{r}$ is already relative to the origin.

Calculating the Moment Vector

The moment vector $\vec{\tau}$ is given by the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$:

$$\vec{\tau} = \vec{r} \times \vec{F}$$

Substituting the given vectors:

$$\vec{\tau} = (\hat{i} + 2\hat{j} + 3\hat{k}) \times (\lambda \hat{k})$$

We can use the distributive property of the cross product and the cross products of the unit vectors:

• $\hat{i} \times \hat{k} = -\hat{j}$
• $\hat{j} \times \hat{k} = \hat{i}$
• $\hat{k} \times \hat{k} = 0$

Now, perform the cross product:

$$\vec{\tau} = (\hat{i} \times \lambda \hat{k}) + (2\hat{j} \times \lambda \hat{k}) + (3\hat{k} \times \lambda \hat{k})$$

$$\vec{\tau} = \lambda (\hat{i} \times \hat{k}) + 2\lambda (\hat{j} \times \hat{k}) + 3\lambda (\hat{k} \times \hat{k})$$

$$\vec{\tau} = \lambda (-\hat{j}) + 2\lambda (\hat{i}) + 3\lambda (0)$$

$$\vec{\tau} = -\lambda \hat{j} + 2\lambda \hat{i}$$

Rearranging the terms in the standard order ($\hat{i}, \hat{j}, \hat{k}$):

$$\vec{\tau} = 2\lambda \hat{i} - \lambda \hat{j} + 0\hat{k}$$

Finding the Magnitude of the Moment

The question asks for the magnitude of the moment. The magnitude of a vector $\vec{V} = V_x \hat{i} + V_y \hat{j} + V_z \hat{k}$ is given by $|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}$.

For the moment vector $\vec{\tau} = 2\lambda \hat{i} - \lambda \hat{j} + 0\hat{k}$, the components are $V_x = 2\lambda$, $V_y = -\lambda$, and $V_z = 0$.

The magnitude of the moment is:

$$|\vec{\tau}| = \sqrt{(2\lambda)^2 + (-\lambda)^2 + (0)^2}$$

$$|\vec{\tau}| = \sqrt{4\lambda^2 + \lambda^2 + 0}$$

$$|\vec{\tau}| = \sqrt{5\lambda^2}$$

Assuming $\lambda$ is a scalar constant, we can write $\sqrt{\lambda^2} = |\lambda|$. However, the options show $\lambda$ outside the square root, suggesting $\lambda$ is taken as a positive scalar or its sign is absorbed. Thus, we consider $\sqrt{5\lambda^2} = \sqrt{5}\sqrt{\lambda^2} = \sqrt{5}|\lambda|$. Given the options, the magnitude is represented as $\sqrt{5}\lambda$.

$$|\vec{\tau}| = \sqrt{5}\lambda$$

Comparing with Options

Let's compare the calculated magnitude with the given options:

• Option 1: $\lambda$
• Option 2: $\sqrt{3}\lambda$
• Option 3: $\sqrt{5}\lambda$
• Option 4: None of the above

Our calculated magnitude is $\sqrt{5}\lambda$, which matches Option 3.

Summary of Calculation Steps

• Identify the position vector $\vec{r}$ and the force vector $\vec{F}$.
• Calculate the cross product $\vec{\tau} = \vec{r} \times \vec{F}$ to find the moment vector.
• Calculate the magnitude of the moment vector $|\vec{\tau}| = \sqrt{\tau_x^2 + \tau_y^2 + \tau_z^2}$.
• Compare the result with the given options.

Revision Table: Key Concepts

Additional Information on Moment of Force

The moment of force is a vector quantity. Its direction is given by the right-hand rule applied to the cross product $\vec{r} \times \vec{F}$. If the fingers of your right hand curl from $\vec{r}$ towards $\vec{F}$, your thumb points in the direction of $\vec{\tau}$. The SI unit for the moment of force is Newton-meter (Nm).

The moment depends on:

• The magnitude of the force, $|\vec{F}|$.
• The distance from the pivot point to the line of action of the force (often called the lever arm), $r_{\perp}$. The magnitude of the moment can also be calculated as $|\vec{\tau}| = r_{\perp} |\vec{F}|$. Alternatively, using the angle $\theta$ between $\vec{r}$ and $\vec{F}$, the magnitude is $|\vec{\tau}| = |\vec{r}| |\vec{F}| \sin(\theta)$. In our vector cross product method, this $\sin(\theta)$ factor is automatically accounted for.
• The direction of the force relative to the position vector.

In three dimensions, the vector cross product is generally the most convenient way to calculate the moment vector and its magnitude.