Question

Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .

Answer 1

Explanation:

Given that,

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

Position of the particle, $r=(3i+4j)\ m$

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

$\tau=F\times r$

$\tau=((-8i)+6j) \times (3i+4j)$

Taking the cross product of above two vectors, we get the value of torque as :

$\tau=(0+0-50k)\ N-m$

(b) Let $\theta$ is the angle between r and F. The angle between two vectors is given by :

$cos\theta=\dfrac{r.F}{|r|.|F|}$

$cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }$

$cos\theta=\dfrac{0}{50}$

$\theta=90^{\circ}$