Question

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What are (a) the torque on the particle about the origin, in unit-vector notation, and (b) the angle between the directions of → r and → F ?

Answer 1

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

Answer 2

Answer:

Explanation:

$\overrightarrow{F}=-8\widehat{i}+6\widehat{j}$

$\overrightarrow{r}=3\widehat{i}+4\widehat{j}$

(a) the formula for the torque is given by

$\overrightarrow{\tau }=\overrightarrow{r} \times \overrightarrow{F}$

$\overrightarrow{\tau }=\left (3\widehat{i}+4\widehat{j} \right )\times \left (-8\widehat{i}+6\widehat{j} \right )$

$\overrightarrow{\tau }=50\widehat{k}$

(b) Let the angle between r and F is θ.

Using the formula

$Cos\theta =\frac{\overrightarrow{F}.\overrightarrow{r}}{\mid \overrightarrow{r}\mid .\mid \overrightarrow{F}\mid }$

$\mid \overrightarrow{r}\mid = \sqrt{3^{2}+4^{2}}=5$

$\mid \overrightarrow{F}\mid = \sqrt{8^{2}+6^{2}}=10$

$\overrightarrow{F}.\overrightarrow{r} = \left ( -8\widehat{i}+6\widehat{j} \right ).\left ( 3\widehat{i}+4\widehat{j} \right )=-24+24=0$

So, Cos θ = 0

θ = 90°

Thus, the angle between f and r is 90°.