Question

Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.

Answer 1

Answer:

$\tau=(-32k)\ N-m$

$\theta=116.55^{\circ}$

Explanation:

Given that.

Force acting on the particle, $F=(-6i + 2.0j)\ N$

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

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

$\tau=F\times r$

$\tau=(-6i + 2.0j)\times (4i+4j)$

$\tau=\begin{pmatrix}0&0&-32\end{pmatrix}$

$\tau=(-32k)\ N-m$

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, $|r|=\sqrt{4^2+4^2}=5.65\ m$

Magnitude of F, $|F|=\sqrt{(-6)^2+2^2}=6.324\ m$

Using dot product formula,

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

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

$cos\theta=\dfrac{-24+8}{6.324\times 5.65}$

$\theta=116.55^{\circ}$

Therefore, this is the required solution.

Answer 2

Answer:

(a) $\overrightarrow{\tau }=32\widehat{K} Nm$

(b)  116.6°

Explanation:

$\overrightarrow{F}=-6\widehat{i}+2\widehat{j}$

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

(a) Torque is defined as

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

$\overrightarrow{\tau }=\left (4\widehat{i}+4\widehat{j} \right )\times \left (-6\widehat{i}+2\widehat{j} \right )$

$\overrightarrow{\tau }=32\widehat{K} Nm$

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

magnitude of r = $\sqrt{4^{2}+4^{2}}=\sqrt{32}$

magnitude of F = $\sqrt{6^{2}+2^{2}}=\sqrt{40}$

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

$Cos\theta =\frac{-24+8}{\sqrt{32\times40}}$

θ = 116.6°