A uniform disk with mass m = 9.42 kg and radius R = 1.31 m lies in the x-y plane and centered at the origin. Three forces act in the +y-direction on the disk: 1) a force 315 N at the edge of the disk on the +x-axis, 2) a force 315 N at the edge of the disk on the –y-axis, and 3) a force 315 N acts at the edge of the disk at an angle θ = 35° above the –x-axis.
3) What is the magnitude of the torque on the disk about the z axis due to F3?
6) What is the z-component of the net torque about the z axis on the disk?
To find the magnitude of the torque on the disk about the z-axis due to F3, we need to calculate the cross product of the position vector and the force vector. The magnitude of the torque can be found using the formula:
τ = r × F
Where τ is the torque, r is the position vector, and F is the force vector.
First, let's find the position vector. The edge of the disk where force F3 acts can be located by using trigonometry. Given the angle θ = 35° above the –x-axis, we can find the x and y components of the position vector r.
x = R * cos(θ)
x = 1.31 m * cos(35°)
x ≈ 1.080 m
y = -R * sin(θ)
y = -1.31 m * sin(35°)
y ≈ -0.746 m
Using the position vector r = (x, y), we can now calculate the cross product with force F3 = (0, 315 N). Since the z-component of the cross product will give us the torque about the z-axis, we only need to consider the z-component of the resulting vector.
r × F3 = (x, y, 0) × (0, 0, 315 N)
= (y * 315 N, -x * 315 N, 0)
= (-233.79 N * m, -340.2 N * m, 0)
The z-component of the result is -340.2 N * m.
Therefore, the z-component of the torque on the disk due to F3 is approximately -340.2 N * m.
torque=force*radius*sinTheta where theta is the angle between position vector and direction of force. T1=315*1.31*sin90 ; T2=315*1.31*sin180; T3=315*1.31*sin135
so for total torque, add them Notice T3 is a subtractive force.