Question

In unit-vector notation, what is the torque about the origin on a jar of jalapeno peppers located at coordinates (3.0m,-2.0m,4.0m) due to (a) force ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{-}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\hat{\mathbf{k}}$,(b) force ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{-}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}\mathbf{+}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\hat{\mathbf{k}}$, (c) the vector sum of ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}_{\mathbf{1}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}_{\mathbf{2}}$? (d) Repeat part (c) for the torque about the point with coordinates (3.0m, 2.0m, 4.0m).

Short answer

1. Torque on the jar due to F₁ is$\left\{\left(6\right)\hat{i}+\left(-3\right)\hat{j}+\left(-6\right)\hat{k}\right\}Nm$
2. Torque on the jar due to F₂ is$\left\{\left(26\right)\hat{i}+\left(3\right)\hat{j}+\left(-18\right)\hat{k}\right\}Nm$
3. Torque on the jar due to F₁ + F₂ is$\left\{\left(32\right)\hat{i}+\left(-24\right)\hat{k}\right\}Nm$
4. Torque about the given point is 0Nm

Step-by-step solution

Identification of given data

Location of jar (3.0m, -2.0m, 4.0m)

$$\begin{gathered} {\stackrel{\rightharpoonup }{F}}_1=\left(3\hat{i}-4\hat{j}+5\hat{k}\right)N \\ {\stackrel{\rightharpoonup }{F}}_2=\left(3\hat{i}-4\hat{j}-5\hat{k}\right)N \end{gathered}$$

To understand the concept of torque

Using the concept of torque, the unknown torque value is calculated. As per the concept, the torque acting on a body is due to the tangential force acting on a body along a radial path of the object in a circular motion. Thus, the cross-vector of the force and radial vector of the object will give the torque value.

Formulae:

The position vector in a 3-D diagram,$\stackrel{\rightharpoonup }{r}=x\hat{i}+y\hat{j}+z\hat{k}$

The force vector in 3-D,$\stackrel{\rightharpoonup }{F}=F_x\hat{i}+F_y\hat{j}+F_z\hat{k}$

The torque acting on the body due to the tangential force,

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\tau }& =& \stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F}\\ & =& \left|\begin{array}{ccc}i& j& k\\ x& y& z\\ F_x& F_y& F_z\end{array}\right|\\ & =& \left(y{F}_z-z{F}_y\right)\hat{i}+\left(z{F}_x-x{F}_z\right)\hat{j}+\left(x{F}_y-y{F}_x\right)\hat{k}\end{array}$

(a) Determining the torque about origin due to F⇀1

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\tau }& =& \stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F}\\ \stackrel{\rightharpoonup }{\tau }& =& \left|\begin{array}{ccc}i& j& k\\ 3& -2& 4\\ 3& -4& 5\end{array}\right|\\ & =& \left(-10+16\right)Nm\hat{i}+\left(12-15\right)Nm\hat{j}+\left(-12+6\right)Nm\hat{k}\\ & =& \left\{\left(6\right)\hat{i}+\left(-3\right)\hat{j}+\left(-6\right)\right\}\hat{k}Nm\end{array}$

(b) Determining the torque about origin due to F⇀2

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\tau }& =& \stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F}\\ \stackrel{\rightharpoonup }{\tau }& =& \left|\begin{array}{ccc}i& j& k\\ 3& -2& 4\\ -3& -4& -5\end{array}\right|\\ & =& \left(10+16\right)Nm\hat{i}+\left(-12+15\right)Nm\hat{j}+\left(-12-6\right)Nm\hat{k}\\ & =& \left\{\left(26\right)\hat{i}+\left(3\right)\hat{j}+\left(18\right)\hat{k}\right\}Nm\end{array}$

(c) Determining the torque about origin due to F⇀1+F⇀2

$\begin{array}{rcl}{\stackrel{\rightharpoonup }{F}}_1+{\stackrel{\rightharpoonup }{F}}_2& =& -8\hat{j}N\\ \stackrel{\rightharpoonup }{\tau }& =& \stackrel{\rightharpoonup }{r}\times \left({\stackrel{\rightharpoonup }{F}}_1+{\stackrel{\rightharpoonup }{F}}_2\right)\\ & =& \left|\begin{array}{ccc}i& j& k\\ 3& -2& 4\\ 0& -8& 0\end{array}\right|\\ & =& \left(32\right)\hat{i}+\left(0\right)\hat{j}+\left(-24+0\right)\hat{k}\\ & =& \left\{\left(32\right)\hat{i}+\left(-24\right)\hat{k}\right\}Nm\end{array}$

(d) Determining the torque about the point with coordinates (3.0m, 2.0m, 4.0m) due to F⇀1+F⇀2

So,$\stackrel{\rightharpoonup }{r}=\stackrel{\rightharpoonup }{r}-\stackrel{\rightharpoonup }{r_0}$

$\stackrel{\rightharpoonup }{r}\text{'}=0\hat{i}-4\hat{j}m+0\hat{k}$

${\stackrel{\rightharpoonup }{F}}_1+{\stackrel{\rightharpoonup }{F}}_2=-8\hat{j}N$

Torque,

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\tau }& =& \stackrel{\rightharpoonup }{r}\times \left({\stackrel{\rightharpoonup }{F}}_1+{\stackrel{\rightharpoonup }{F}}_2\right)\end{array}$

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\tau }& =& \stackrel{\rightharpoonup }{r}\times \left({\stackrel{\rightharpoonup }{F}}_1+{\stackrel{\rightharpoonup }{F}}_2\right)\\ & =& \left|\begin{array}{ccc}i& j& k\\ 0& -4& 0\\ 0& -8& 0\end{array}\right|\\ & =& \left(0\right)\hat{i}Nm+\left(0\right)\hat{j}Nm+\left(0\right)\hat{k}Nm\\ & =& 0Nm\end{array}$