Respuesta :
Answer with Explanation:
We are given that
[tex]F=-8\hat{i}+6\hat{j}[/tex]
[tex]r=3\hat{i}+4\hat{j}[/tex]
a.We have to find the torque on the particle about the origin.
We know that
Torque=[tex]\tau=r\times F=\begin{vmatrix}i&j&k\\3&4&0\\-8&6&0\end{vmatrix}[/tex]
By using the formula
[tex]\tau=50\hat{k}[/tex]
b.[tex]\mid \tau\mid =\mid F\mid \mid r\mid sin\theta[/tex]
[tex]\mid F\mid=\sqrt{(-8)^2+(6)^2}=10[/tex]
[tex]\mid r\mid=\sqrt{3^2+4^2}=5[/tex]
[tex]\mid \tau\mid=\sqrt{(-50)^2}=50[/tex]
Substitute the values then we get
[tex]50=10\times 5 sin\theta[/tex]
[tex]sin\theta=\frac{50}{50}=1[/tex]
[tex]sin\theta=sin90^{\circ}[/tex]
Because [tex]sin90^{\circ}=1[/tex]
[tex]\theta=90^{\circ}[/tex]
(a) the torque is a vector product of force and position vector acting on the particle and its magnitude is 50 Nm
(b) the angle between force and position vector is 90°
Vector product:
(a) Given that the force acting on the particle is :
F = -8N i + 6N j
and the position vector of the particle is:
r = 3m i + 4m j
Noe, the torque is the cross product of force and position vector of the particle, which is given by:
[tex]\tau=r\times F= \begin{vmatrix}i &j&k \\3 & 4&0\\-8&6&0\end{vmatrix}\\\\\tau=50k[/tex]
(b) also, the vector product is given by:
[tex]\tau=rFsin\theta[/tex]
the magnitudes are;
|τ| = 50 Nm
[tex]|F|=\sqrt{(-8)^2+6^2}=10 \;N[/tex]
and
[tex]|r|=\sqrt{3^2+4^2}=5\;m[/tex]
so
[tex]\tau=rFsin\theta\\\\50=5\times10\times sin\theta\\\\sin\theta=1\\\\\theta=90^o[/tex]
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