A uniform disk with mass m = 8.83 kg and radius R = 1.44 m lies in the x-y plane

Question

A uniform disk with mass m = 8.83 kg and radius R = 1.44 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 326 N at the edge of the disk on the +x-axis, 2) a force 326 N at the edge of the disk on the -y-axis, and 3) a force 326 N acts at the edge of the disk at an angle 0 = 31 degree above the -x-axis. What is the magnitude of the torque on the disk about the z axis due to F_1? What is the magnitude of the torque on the disk about the z axis due to F_2? What is the magnitude of the torque on the disk about the z axis due to F_3? What is the x-component of the net torque about the z axis on the disk?

Explanation / Answer

drop a perpendicular from the centre of the disk to the force line F3.

let that distance be D.

then as given that angle with -ve x axis is 31 degrees,

cos(theta)=D/1.44

==>D=1.2343 m

y coordinate of the point where F3 touches the disk, be d.

then sin(theta)=d/1.44

==>d=.74165 m

hence in vector notation, point of contact of the force from the origin/z axis=-1.2343 i + 0.64165 j

torque=cross product of distance vector and force

=(-1.2343 i + 0.74165 j )*(326 j)

=-402.3818 k

hence the torque is acting along -ve z axis and with a magnitude of 402.3818 N.m

hence answers are:

3. 402.3818 N.m

Q4. 0

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