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 ) ( ) . 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

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

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

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

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

$\tau = 32k +18k$

$\tau = 50 k$

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

$cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}$

$cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}$

$cos\theta = -0.24$

$\theta = cos^{-1} (-0.24)$

$\theta = 103.88\°$

Therefore the angle between the ratio and the force is 103.88°