A couple is defined as two parallel forces, separated by a distance, that have equal magnitudes but opposite directions. A couple only produces a rotation in a specified direction. The moment produced by a couple is called a couple moment.
A submarine hatch door is to be opened by applying two oppositely oriented forces of equal magnitude F=265N at points A and B on the hatch door wheel. The radii of the wheel's inner and outer rings are r1 = 0.470 m and r2 = 0.200 m, respectively.
Calculate the moments MA and MB about point D for the forces applied at points A and B. Then, determine the resulting couple moment MR. Assume that a positive moment produces a counterclockwise rotation whereas a negative moment produces a clockwise rotation.

MA = 178 Nm

MB = 72 Nm

MR = 249 Nm

For the moments MA the lever arm was taken as r1 + r2 which is the distance of the point of application of the force at point A to point D.

For the moment MB, the distance of the point of application of force is r1 - r2 which is the distance from the outer ring to the inner ring.

The couple moment is given by F × r1. Which is basically a sum of the moments of both forces applied on the wheel.

Explanation:

See the attachment for detail of the calculation.

You have submitted a wrong question, because the radi of the inner ring cannot be greater than the radi of the outer ring. Please let me assume the the radi of the inner ring to be the value for the outer ring, and the radi of the outer ring to be the value of the inner ring. That is;

Outer ring (r1) = 0.470m

Inner ring (r2) = 0.200m

ANSWER:

MA = 124.55Nm

MB = 53Nm

MR = 177.55Nm

EXPLANATION: Please see image attached to understand how the question is been analysed before proceeding with this explanations.

Using Archimedes principle

M = rF......................(1)

M is moment

r is the distance of the force to the center point

F is the force.

FIND THE MOMENT OF FORCE APPLIED AT POINT A (MA)

Because point A is the first point for the force which rotates the ring and act outwardly to the position of the applied force. Therefore the force at point A is on the outer ring

r1 = 0.470m

F = 265N

Using equation 1

MA = 265 × 0.470 = 124.55Nm

FIND THE MOMENT OF FORCE APPLIED AT POINT B (MB)

Because point B is the second point for the force which rotates the ring and act inwardly to the position of the applied force. Therefore the force at point B is on the inner ring.

r2 = 0.200m

F = 265N

Using equation 1

MB = 0.200 × 265 = 53Nm

THE RESULTING COUPLE MOMENT

The resulting moment is the sum of the moment acting on point A and B

MR = MA + MB

MR = 124.55 + 53 = 177.55Nm

increases it

resonant frequency