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Chapter 12: Problem 44

Answer the following questions about torque. A pump handle has a pivot at (0,0,0) and extends to \(P(5,0,-5)\) A force \(\mathbf{F}=\langle 1,0,-10\rangle\) is applied at \(P .\) Find the magnitude and direction of the torque about the pivot.

Short Answer

Expert verified
Answer: The magnitude of the torque is 45, and its direction is along the positive y-axis, given by the vector \(\langle 0,1,0\rangle\).

Step by step solution

01

Finding the position vector

First, we need to find the position vector \(\mathbf{r}\), which points from the pivot point to the point where the force is applied. The position vector is given by the coordinates of point P(5,0,-5): \(\mathbf{r}=\langle 5,0,-5\rangle\).
02

Calculating the torque

Now, we'll calculate the torque \(\mathbf{T}\) about the pivot point using the vector product (cross product) between the position vector \(\mathbf{r}\) and the force vector \(\mathbf{F}\): $$\mathbf{T}=\mathbf{r}\times\mathbf{F}$$ Here, \(\mathbf{r}=\langle 5,0,-5\rangle\) and \(\mathbf{F}=\langle 1,0,-10\rangle\).
03

Computing the cross product

To compute the cross product, we will use the formula: \(\mathbf{T}=\langle T_x,T_y,T_z \rangle\) where \(T_x=(0)(-10)-(0)(-5)\), \(T_y=(-5)(-10)-(5)(1)\), and \(T_z=(5)(0)-(1)(0)\). Computing these components gives: \(T_x=0\), \(T_y=45\), and \(T_z=0\). So, the torque vector is \(\mathbf{T}=\langle 0,45,0\rangle\).
04

Finding the magnitude of torque

Now, we'll compute the magnitude of the torque vector \(\mathbf{T}\) using the formula: $$|\mathbf{T}|=\sqrt{T_x^2+T_y^2+T_z^2}$$ Substituting the previously computed components, we have: $$|\mathbf{T}|=\sqrt{0^2+45^2+0^2}=\sqrt{2025}=45$$ The magnitude of the torque is 45.
05

Finding the direction of torque

To find the direction of torque, we must normalize the torque vector. We can normalize the vector by dividing each component by the magnitude of torque. $$\mathbf{T}_{norm}=\frac{\mathbf{T}}{|\mathbf{T}|}=\frac{\langle 0,45,0\rangle}{45}=\langle 0,1,0\rangle$$ The direction of the torque is given by the normalized torque vector, which is \(\langle 0,1,0\rangle\), pointing in the positive y-direction. To conclude, the magnitude of the torque about the pivot is 45, and its direction is along the positive y-axis, given by the vector \(\langle 0,1,0\rangle\).

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Most popular questions from this chapter

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