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olchik [2.2K]
3 years ago
12

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 → ?
Physics
1 answer:
disa [49]3 years ago
3 0

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}

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