Answer
(a) $r = (3.00~\hat{i}+4.00~\hat{j})~m$
(b) The magnitude of the torque is 37.0 N m, and it is directed in the clockwise direction.
Work Step by Step
(a) x = 3.00 m and y = 4.00 m
$r = (3.00~\hat{i}+4.00~\hat{j})~m$
(b) $F = (7.00~\hat{i}-3.00~\hat{j})~N$
We can find the magnitude of the force vector.
$F = \sqrt{(7.00~N)^2+(-3.00~N)^2}$
$F = 7.62~N$
We can find the angle below the positive x-axis that the force vector is applied.
$tan(\theta) = \frac{3.00}{7.00}$
$\theta = arctan(\frac{3.00}{7.00})$
$\theta = 23.2^{\circ}$
We can find the magnitude of the position vector.
$r = \sqrt{(3.00~m)^2+(4.00~m)^2}$
$r = 5.00~m$
We can find the angle above the positive x-axis of the position vector.
$tan(\theta) = \frac{4.00}{3.00}$
$\theta = arctan(\frac{4.00}{3.00})$
$\theta = 53.1^{\circ}$
The angle between the position vector and the force vector is $23.2^{\circ}+53.1^{\circ}$, which is $76.3^{\circ}$
$\tau = r\times F$
$\tau = -r~ F~sin(76.3^{\circ})$
$\tau = -(5.00~m)(7.62~N)~sin(76.3^{\circ})$
$\tau = -37.0~N~m$
The magnitude of the torque is 37.0 N m, and it is directed in the clockwise direction.
