1) τ = 437 N m, 2) τ = 437 N m
, 3) τ = 344.4 N m
, 7) α = 142.5 rad/s² and

8) Em = 1.95 105 J

Explanation:

The torque is given by the expression

τ = F x R

Bold indicates vectors, we can calculate the magnitude of this torque with the expression

τ = F R sin θ

The direction is given by perpendicular to the two vectors F and R

1) To calculate the torque due to the F1 force. The angle between F1 and R is 90°

sin 90 = 1

τ = F R

τ = 319 1.37

τ = 437 N m

In the positive Z direction

2) The torque due to F2 again the angle is 90 ° without 90 = 1

τ = F R

τ = (-319) 1.37 sin 270

τ = 437 N m

In the positive z direction

3) The torque F3 where θ = 32 ° measured from the negative side of the x axis

τ = F R sin θ

Angle is measured from the positive side of the x-axis, so this angle is

θ = 180 - 32 = 128 °

τ = 319 1.37 sin 128

τ = 344.4 N m

4) The x components of this torque is

F1 R + F3 R cos 128 = 437 + 319 1.37 cos 128 = 437 + 269

τₓ = 706 N m

5) the component and

F2 R +F3 R sin 128 = 437 + 344.4

τ y = 781.4 N m

The component has to be perpendicular to the moment

7) angular acceleration

τ = I α

The moment of inertia of a disk is

I = ½ M R²

I = ½ 9.11 1.37²

I = 8.55 Kg m²

α = τ / I

α = (τ1 + τ2 + τ3) / I

α = (437 + 437 + 344.4) / 8.55

α = 142.5 rad / s²

8) the disk starts at rest, let's calculate the speed in the given time

w = w₀ + α t

w = 0 + 142.5 1.5

w = 213.75 rad / s

Energy is

Em = K = ½ I w²

Em = ½ 8.55 213.75²

Em = 1.95 105 J