1. A circular disc is rotated about an axis O through its center by the application of two forces. A force of

Question

1. A circular disc is rotated about an axis O through its center by the application of two forces. A force of magnitude 11 N is exerted at a distance of 0.34 m from the axis and at an angle of 58° from a radial line extending from the axis through the point of application Q of the force. A second force of magnitude 15 N is exerted at a distance of 0.26 m from the axis and at an angle of 119° from a radial line extending from the axis through the point of application P of the force. Determine the net torque on the disc about its center and which way the net torque accelerates the disc.

Answer 1

Yes, the answer is cats.

Answer 2

For these angles, I'm assuming we're using the right hand rule, meaning the angles are measured counterclockwise from the outwards radial direction. The formula for torque is T = F^ x R^, where F^ is the force vector, R^ is the radius vector, and x signifies the cross product. This is equal to FR*sin(theta), where F and R are the magnitudes of F^ and R^, respectively, and theta is the angle between them. Since you have been given F, R, and theta for both forces, you just need to apply this formula twice and sum the results. If you get a positive result, the disc accelerates in the counterclockwise direction (due to the right hand rule); a negative result means clockwise.

Answer 3

The component of the 11 N force perpendicular to the radius is 11 sin 58.

The torque of this force about the axis is force x distance. = 11 sin 58 x 0.34 =3.17 Nm.

The component of the 15 N force perpendicular to the radius is 15 sin 119.

The torque of this force about the axis is force x distance. = 15 sin 119 x 0.26 =3.41Nm.

The net torque is 3.17 + 3.41 = 6.58 Nm.

Since angle is measured anticlockwise, the toque produces rotation in the anticlockwise direction.