Respuesta :
Answer:
[tex]\tau=(-32k)\ N-m[/tex]
[tex]\theta=116.55^{\circ}[/tex]
Explanation:
Given that.
Force acting on the particle, [tex]F=(-6i + 2.0j)\ N[/tex]
Position of the particle, [tex]r=(4i+4j)\ m[/tex]
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 :
[tex]\tau=F\times r[/tex]
[tex]\tau=(-6i + 2.0j)\times (4i+4j)[/tex]
[tex]\tau=\begin{pmatrix}0&0&-32\end{pmatrix}[/tex]
[tex]\tau=(-32k)\ N-m[/tex]
So, the torque on the particle about the origin is (32 N-m).
(b) Magnitude of r, [tex]|r|=\sqrt{4^2+4^2}=5.65\ m[/tex]
Magnitude of F, [tex]|F|=\sqrt{(-6)^2+2^2}=6.324\ m[/tex]
Using dot product formula,
[tex]F{\circ}\ r=|F|.|r|\ cos\theta[/tex]
[tex]cos\theta=\dfrac{F{\circ} r}{|F|.|r|}[/tex]
[tex]cos\theta=\dfrac{-24+8}{6.324\times 5.65}[/tex]
[tex]\theta=116.55^{\circ}[/tex]
Therefore, this is the required solution.
Answer:
(a) [tex]\overrightarrow{\tau }=32\widehat{K} Nm[/tex]
(b) 116.6°
Explanation:
[tex]\overrightarrow{F}=-6\widehat{i}+2\widehat{j}[/tex]
[tex]\overrightarrow{r}=4\widehat{i}+4\widehat{j}[/tex]
(a) Torque is defined as
[tex]\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}[/tex]
[tex]\overrightarrow{\tau }=\left (4\widehat{i}+4\widehat{j} \right )\times \left (-6\widehat{i}+2\widehat{j} \right )[/tex]
[tex]\overrightarrow{\tau }=32\widehat{K} Nm[/tex]
(b) Let θ be the angle between the r and F.
magnitude of r = [tex]\sqrt{4^{2}+4^{2}}=\sqrt{32}[/tex]
magnitude of F = [tex]\sqrt{6^{2}+2^{2}}=\sqrt{40}[/tex]
[tex]Cos\theta =\frac{\overrightarrow{r}.\overrightarrow{F}}{rF}[/tex]
[tex]Cos\theta =\frac{-24+8}{\sqrt{32\times40}}[/tex]
θ = 116.6°
