Question

Force$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{N}\mathbf{)}\hat{\mathbf{j}}$acts on a particle with position vector$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{=}\left(3.0m\right)\hat{\mathbf{i}}\mathbf{+}\left(4.0m\right)\hat{\mathbf{j}}$. (a) What is the torque on the particle about the origin, in unit-vector notation? (b) What is the angle between the directions of$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}$ and$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$?

Short answer

1. Torque on particle about the origin in unit vector notation is$\left(50\right)\hat{k}Nm$
2. Angle between directions of$\stackrel{\rightharpoonup }{r}and\stackrel{\rightharpoonup }{F},\varphi ={90}^{\circ }$

Step-by-step solution

Identification of given data

$$\begin{gathered} \stackrel{\rightharpoonup }{F}=\left(-8\hat{i}+6\hat{j}+0\hat{k}\right)N \\ \stackrel{\rightharpoonup }{r}=\left(3\hat{i}+4\hat{j}+0\hat{k}\right)m \end{gathered}$$

To understand the concept of torque and angle

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,

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

$\left|\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F}\right|=rF\sin \varphi$ …(i)

$r=\sqrt{x^2+y^2}andF=\sqrt{F_x^2+F_y^2}$

(a) Determining the torque on the particle about the origin

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\tau }& =& \stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F}\\ \stackrel{\rightharpoonup }{\tau }& =& \left|\begin{array}{ccc}i& j& k\\ 3& 4& 0\\ -8& 6& 0\end{array}\right|\\ & =& \left(0\right)\hat{i}Nm+\left(0\right)\hat{j}Nm+\left(18+32\right)\hat{k}Nm\\ & =& \left(0\right)\hat{i}+\left(0\right)\hat{j}+\left(50\right)\hat{k}Nm\end{array}$

(b) Determining the angle between the directions of r⇀ and F⇀

We know,

$\begin{array}{rcl}r& =& \sqrt{x^2+y^2}\\ & =& \sqrt{64m^2+36m^2}\\ & =& 10m\\ F& =& \sqrt{F_x^2+F_y^2}\\ & =& \left(\sqrt{9+6}\right)N\\ & =& 5N\\ \tau & =& rF\\ & =& 50Nm\end{array}$

$\left|\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F}\right|=50Nm$

Using equation (i)

$\begin{array}{rcl}\left|\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F}\right|& =& rF\sin \varphi \\ 50Nm& =& 50Nm\times \sin \varphi \\ \sin \varphi & =& 1\\ \varphi & =& {90}^o\end{array}$

The angle between the directions of $\stackrel{\rightharpoonup }{r}$and $\stackrel{\rightharpoonup }{F}$ is$\begin{array}{rcl}\varphi & =& {90}^o\end{array}$