Question

Force F → = (−8.0 N)iˆ + (6.0 N)jˆ acts on a particle with position vector r → = (3.0 m)iˆ + (4.0 m)jˆ. 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

Answer with Explanation:

We are given that

$F=-8\hat{i}+6\hat{j}$

$r=3\hat{i}+4\hat{j}$

a.We have to find the torque on the particle about the origin.

We know that

Torque=$\tau=r\times F=\begin{vmatrix}i&j&k\\3&4&0\\-8&6&0\end{vmatrix}$

By using the formula

$\tau=50\hat{k}$

b.$\mid \tau\mid =\mid F\mid \mid r\mid sin\theta$

$\mid F\mid=\sqrt{(-8)^2+(6)^2}=10$

$\mid r\mid=\sqrt{3^2+4^2}=5$

$\mid \tau\mid=\sqrt{(-50)^2}=50$

Substitute the values then we get

$50=10\times 5 sin\theta$

$sin\theta=\frac{50}{50}=1$

$sin\theta=sin90^{\circ}$

Because $sin90^{\circ}=1$

$\theta=90^{\circ}$