Question

Force $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\left(2.0N\right)\hat{\mathbf{i}}\mathbf{-}\left(3.0N\right)\hat{\mathbf{k}}$acts on a pebble with position vector$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{=}\left(0.50m\right)\hat{\mathbf{j}}\mathbf{-}\left(2.0m\right)\hat{\mathbf{k}}$relative to the origin. In unit- vector notation, (a) What is the resulting torque on the pebble about the origin and (b) What is the resulting torque on the pebble about the point$\left(2.0m,0,-3.0m\right)\mathbf{?}$

Short answer

1. Net torque at origin is$\left\{\left(-1.5\right)\hat{i}+\left(-4\right)\hat{j}+\left(-1\right)\hat{k}\right\}Nm$
2. Net torque at given point$\left\{\left(-1.5\right)\hat{i}+\left(-4\right)\hat{j}+\left(-1\right)\hat{k}\right\}Nm$

Step-by-step solution

Identification of given data

$$\begin{gathered} \stackrel{\rightharpoonup }{r}=\left(0m\right)\hat{i}+\left(0.50m\right)\hat{j}-\left(2.0m\right)\hat{k} \\ \stackrel{\rightharpoonup }{F}=2N\hat{i}+0\hat{j}-3N\hat{k} \\ \stackrel{\rightharpoonup }{r}=\left(2m\right)\hat{i}+\left(0m\right)\hat{j}-\left(3.0m\right)\hat{k} \end{gathered}$$

To understand the concept of torque and force

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. Hence, through comparison of the torque given and calculated format, the force value can be given.

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 resulting torque on the pebble about the origin

Using the formula for the torque, we have,

$\begin{array}{rcl}\hat{\tau }& =& \left|\begin{array}{ccc}i& j& k\\ 0& 0.5& -2\\ 2& 0& -3\end{array}\right|\\ & =& \left\{\left(-1.5\right)\hat{i}+\left(-4\right)\hat{j}+\left(-1\right)\hat{k}\right\}Nm\end{array}$

(b) Determining the resulting torque on the pebble about the point (2.0m, -3.0m)

Now,

$\begin{array}{rcl}\stackrel{\rightharpoonup }{r}\text{'}& =& \stackrel{\rightharpoonup }{r}-{\stackrel{\rightharpoonup }{r}}_0\\ & =& \left(-2.0\hat{i}+0.5\hat{j}+1\hat{k}\right)m\\ \stackrel{\rightharpoonup }{F}& =& 2N\hat{i}+0\hat{j}-3N\hat{k}\\ \stackrel{\rightharpoonup }{\tau }& =& \left|\begin{array}{ccc}i& j& k\\ -2& 0.5& 1\\ 2& 0& -3\end{array}\right|\\ & =& \left(-1.5\right)Nm\hat{i}+\left(-6+2\right)Nm\hat{j}+\left(-2\times 0-2\times 0.5\right)Nm\hat{k}\\ & =& \left\{\left(-1.5\right)\hat{i}+\left(-4\right)\hat{j}+\left(-1\right)\hat{k}\right\}Nm\end{array}$