A car mechanic applies a force of 800 N to a wrench to loosen a bolt. She applies the force perpendicular to the arm of the wrench. The distance from the bolt to her hand is 0.40 m... A car mechanic applies a force of 800 N to a wrench to loosen a bolt. She applies the force perpendicular to the arm of the wrench. The distance from the bolt to her hand is 0.40 m. What is the magnitude of the torque applied? Read more

Steps to Solve

1. Find the vectors $\vec{AB}$ and $\vec{AC}$

To find the area of the triangle, we first need to find two vectors that represent two sides of the triangle. We can use the points $A$, $B$, and $C$ to find these vectors:

$\vec{AB} = B - A = (1 - 3, -1 - (-1), -3 - 2) = (-2, 0, -5)$

$\vec{AC} = C - A = (4 - 3, -3 - (-1), 1 - 2) = (1, -2, -1)$

2. Calculate the cross product of $\vec{AB}$ and $\vec{AC}$

The area of the triangle is half the magnitude of the cross product of the vectors $\vec{AB}$ and $\vec{AC}$. First, compute the cross product:

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 0 & -5 \ 1 & -2 & -1 \end{vmatrix} = \hat{i}(0 \cdot (-1) - (-5) \cdot (-2)) - \hat{j}((-2) \cdot (-1) - (-5) \cdot 1) + \hat{k}((-2) \cdot (-2) - 0 \cdot 1) = -10\hat{i} - 7\hat{j} + 4\hat{k} = (-10, -7, 4)$

3. Find the magnitude of the cross product

Now, we calculate the magnitude of the cross product vector:

$|\vec{AB} \times \vec{AC}| = \sqrt{(-10)^2 + (-7)^2 + 4^2} = \sqrt{100 + 49 + 16} = \sqrt{165}$

4. Calculate the area of the triangle

The area of the triangle is half the magnitude of the cross product:

Area $= \frac{1}{2}|\vec{AB} \times \vec{AC}| = \frac{1}{2}\sqrt{165}$

5. Find the area of the quadrilateral

To find the area of the quadrilateral $ABCD$, we can divide it into two triangles, $ABD$ and $BCD$, and find the area of each triangle separately, then sum them.

6. Area of triangle $ABD$

$\vec{AB} = B - A = (3-1, 5-0) = (2, 5)$ $\vec{AD} = D - A = (5-1, -2-0) = (4, -2)$ Area of $ABD = \frac{1}{2} |(2)(-2) - (5)(4)| = \frac{1}{2} |-4 - 20| = \frac{1}{2} |-24| = 12$

7. Area of triangle $BCD$

$\vec{BC} = C - B = (7-3, 2-5) = (4, -3)$ $\vec{BD} = D - B = (5-3, -2-5) = (2, -7)$ Area of $BCD = \frac{1}{2} |(4)(-7) - (-3)(2)| = \frac{1}{2} |-28 + 6| = \frac{1}{2} |-22| = 11$

8. Add the areas of the triangles

Area of quadrilateral $ABCD$ = Area of $ABD$ + Area of $BCD$ = $12 + 11 = 23$.

9. Calculate the torque

To calculate the torque, we use the formula $\tau = rF\sin(\theta)$, where $r$ is the distance from the pivot point, $F$ is the applied force, and $\theta$ is the angle between the force and the wrench arm. Since the force is applied perpendicular to the wrench arm, $\theta = 90^\circ$, and $\sin(90^\circ) = 1$. Thus, $\tau = rF$. $\tau = (0.40 , \text{m})(800 , \text{N}) = 320 , \text{N} \cdot \text{m}$