Respuesta :
Explanation:
It is given that,
Force, [tex]F=(-8\ N)i+(6\ N)j[/tex]
Position vector, [tex]r=(3i+4j)\ m[/tex]
(a) The torque on the particle about the origin is given by :
[tex]\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m[/tex]
(b) To find the angle between r and F use dot product formula as :
[tex]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}[/tex]
Hence, this is the required solution.
Answer:
Explanation:
[tex]\overrightarrow{F}=-8\widehat{i}+6\widehat{j}[/tex]
[tex]\overrightarrow{r}=3\widehat{i}+4\widehat{j}[/tex]
(a) the formula for the torque is given by
[tex]\overrightarrow{\tau }=\overrightarrow{r} \times \overrightarrow{F}[/tex]
[tex]\overrightarrow{\tau }=\left (3\widehat{i}+4\widehat{j} \right )\times \left (-8\widehat{i}+6\widehat{j} \right )[/tex]
[tex]\overrightarrow{\tau }=50\widehat{k}[/tex]
(b) Let the angle between r and F is θ.
Using the formula
[tex]Cos\theta =\frac{\overrightarrow{F}.\overrightarrow{r}}{\mid \overrightarrow{r}\mid .\mid \overrightarrow{F}\mid }[/tex]
[tex]\mid \overrightarrow{r}\mid = \sqrt{3^{2}+4^{2}}=5[/tex]
[tex]\mid \overrightarrow{F}\mid = \sqrt{8^{2}+6^{2}}=10[/tex]
[tex]\overrightarrow{F}.\overrightarrow{r} = \left ( -8\widehat{i}+6\widehat{j} \right ).\left ( 3\widehat{i}+4\widehat{j} \right )=-24+24=0[/tex]
So, Cos θ = 0
θ = 90°
Thus, the angle between f and r is 90°.
