What is the moment of the couple formed by the two forces applied to the corners b and d of a rectangular plate, given that p = 88-n? Resolve each force into horizontal and vertical components, and add the moments of the resulting couples?

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Answer 1

Answer:

The moment of the couple formed by the two forces applied to the corners b and d of a rectangular plate, given that p = 88-n, is 7.33 N * m.

What is the moment of the couple formed by the two forces applied to the corners b and d of a rectangular plate?

To resolve each force into horizontal and vertical components, we can use the following equations:

Horizontal component = P * cos(theta)

Vertical component = P * sin(theta)

where P is the magnitude of the force and theta is the angle between the force and the horizontal.

In this case, both forces are applied at a 50-degree angle to the horizontal. Therefore, the horizontal and vertical components of each force are:

Horizontal component = 88 N * cos(50 degrees) = 51.42 N

Vertical component = 88 N * sin(50 degrees) = 61.28 N

To calculate the moment of the couple, we need to multiply the force by the perpendicular distance between the line of action of the force and the origin of the moment. In this case, the perpendicular distance is the length of the plate.

The moment of the couple formed by the two forces is:

Moment = (51.42 N * 0.5 m) + (61.28 N * 0.3 m) = 7.33 N * m

Therefore, the moment of the couple formed by the two forces applied to the corners b and d of a rectangular plate, given that p = 88-n, is 7.33 N * m.

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