To solve the problem, we need to compute the length of the lever arm $L$ required to produce a torque of magnitude 300 ft-lb, given that the force has a magnitude of 100 pounds and points in the direction of the vector $\mathbf{V} = (10, 0, 16)$.

Step 1: Recall the formula for torque

The torque $\mathbf{\tau}$ is calculated using the following formula: $\mathbf{\tau} = \mathbf{r} \times \mathbf{F}$ Where:

• $\mathbf{\tau}$ is the torque vector.
• $\mathbf{r} = (r_0, 0, 0)$ is the position vector, which represents the lever arm.
• $\mathbf{F} = (F_x, F_y, F_z) = (10, 0, 16)$ is the force vector (direction of the force).
• $\times$ denotes the cross product.

The magnitude of the torque is given by: $|\mathbf{\tau}| = |\mathbf{r}| |\mathbf{F}| \sin \theta$ Where:

• $|\mathbf{\tau}| = 300$ ft-lb is the magnitude of the torque.
• $|\mathbf{r}| = L$ is the length of the lever arm (which we need to find).
• $|\mathbf{F}| = 100$ pounds is the magnitude of the force.
• $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{F}$.

Step 2: Compute the angle between $\mathbf{r}$ and $\mathbf{F}$

Since $\mathbf{r} = (r_0, 0, 0)$, the force vector is at an angle with $\mathbf{r}$ along the x-axis. To compute the sine of the angle $\theta$, we will use the following relation: $\sin \theta = \frac{|\mathbf{r} \times \mathbf{F}|}{|\mathbf{r}| |\mathbf{F}|}$

First, compute the cross product $\mathbf{r} \times \mathbf{F}$:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ r_0 & 0 & 0 \\ 10 & 0 & 16 \end{vmatrix}$$ Evaluating the determinant: $$\mathbf{r} \times \mathbf{F} = \mathbf{i}(0 \cdot 16 - 0 \cdot 0) - \mathbf{j}(r_0 \cdot 16 - 0 \cdot 10) + \mathbf{k}(r_0 \cdot 0 - 0 \cdot 10)$$ $$\mathbf{r} \times \mathbf{F} = -16 r_0 \mathbf{j}$$ Thus, the magnitude of the cross product is: $$|\mathbf{r} \times \mathbf{F}| = |16 r_0|$$ ### Step 3: Set up the torque equation Now we can set up the torque equation: $$|\mathbf{\tau}| = |\mathbf{r}| |\mathbf{F}| \sin \theta = |r_0| \cdot 100 \cdot \frac{|16 r_0|}{|r_0| \cdot 100}$$ Simplifying: $$300 = |r_0| \cdot 16$$ $$r_0 = \frac{300}{16} = 18.75 \text{ feet}$$ Thus, the length of the lever arm $$L = r_0 = 18.75$$ feet. ### Final Answer: The length of the lever arm required to produce a torque of 300 ft-lb is $$L = 18.75$$ feet. --- Would you like more details or have any further questions? Here are 5 related questions: 1. How is the direction of the torque determined from the cross product? 2. What other factors influence the torque produced by a force? 3. Can the angle between the force and lever arm be used to reduce torque? 4. How does changing the magnitude of force affect the required lever arm length? 5. What if the force was applied at an angle—how would that change the torque? **Tip:** Torque is maximized when the force is applied perpendicular to the lever arm, as $$\sin 90^\circ = 1$$.