Finding the Moment of a Force about a Point

In summary, the task is to find the moment of point A about point D. The equation to use is MD= rA/D X F, where r is the radius vector and F is the force vector. The trig function needed can be found by taking the cross product of the two vectors. By breaking the vectors into their components and using the cross product, the task can be solved easily.
  • #1
sdoug041
26
0

Homework Statement



I need to find the moment of A about D. We are given the Fg on the object to be 49N and the distance from A to D is 27.5cm. I've attached a drawing to help demonstrate.


Homework Equations



MD= rA/D X F

The Attempt at a Solution



I know the radius between A and D is 27.5cm but I'm not sure how to find out how much of the force is actually acting on D! I know I need to use sin or cos of a deduced angle (15 degrees or one relating to that). I'm not sure which trig operation or angle to use.

Would anyone be able to shed some light on how to go about? Much thanks.
 

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  • #2
sdoug041 said:

Homework Statement



I need to find the moment of A about D. We are given the Fg on the object to be 49N and the distance from A to D is 27.5cm. I've attached a drawing to help demonstrate.


Homework Equations



MD= rA/D X F

The Attempt at a Solution



I know the radius between A and D is 27.5cm but I'm not sure how to find out how much of the force is actually acting on D! I know I need to use sin or cos of a deduced angle (15 degrees or one relating to that). I'm not sure which trig operation or angle to use.

Would anyone be able to shed some light on how to go about? Much thanks.

Welcome to the PF. The equation that you wrote is a vector equation, so you only need to write the vectors r and F, and take their cross product. The definition of the cross product includes the trig function that you need to use. Does that help?
 
  • #3
Muchly appreciated I understand right now. Simply break the radius and force vector into it's components and cross. I knew it was easier than I thought; my frustration was unjustified!

Thanks man.
 

1. What is a moment about a point problem?

A moment about a point problem is a physics concept that involves calculating the turning effect or torque of an object around a fixed point. It is used to determine the rotational motion of an object in response to an applied force.

2. How is moment about a point different from moment about an axis?

The difference between moment about a point and moment about an axis lies in the point of reference. Moment about a point is calculated with respect to a fixed point, while moment about an axis is calculated with respect to a fixed axis of rotation. The two concepts are related, but they are not interchangeable.

3. What factors affect the moment about a point?

The moment about a point is affected by the magnitude and direction of the applied force, as well as the distance between the point of rotation and the force. The direction of the force relative to the point of rotation also plays a role in determining the moment.

4. How is moment about a point calculated?

The moment about a point is calculated by multiplying the magnitude of the force by the perpendicular distance between the point of rotation and the line of action of the force. This can be represented by the equation M = F * d, where M is the moment, F is the force, and d is the distance.

5. What are some real-life applications of moment about a point?

Moment about a point is used in engineering and construction to design and analyze structures such as bridges and buildings. It is also used in sports, such as in the mechanics of throwing a ball or swinging a bat. Additionally, it is important in understanding the movement of objects in nature, such as the rotation of planets and stars.

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