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# M2 (Edexcel) - Exercise 5B, Q6 - Moments help watch

1. Hi guys,

I'm stuck on Q6 of Exercise 5B in the M2 Edexcel Heinemann book.

From the solution bank:

M(A) : 5g x 1 = (F X 1) + (F X 2)

I understand where (F X 2) comes from, but, taking moments about A, how did they get (F X 1) and (5g x 1)?

Many thanks.
2. You take Moments About A ( top left corner of the shape), assuming your diagram is correct
Clockwise Moments are: 5g x 1m ( weight of the shape is a force and its distance to A is half way between AD because its the centre of mass and because AD is 2m, then distance From A to 5g force is 1m)

Anticlockwise moments are: F x 2m ( that force at the top pulling it up) + F x 1m ( that force at the bottom pulling corner C to the right, if you look closely you can see that the force is already perrpedicular to A and distance to this force is AB )

M(A) 5g x 1 = Fx2 + Fx1

Hope It helped, Im self-studying m2
3. (Original post by ms94uk)
You take Moments About A ( top left corner of the shape), assuming your diagram is correct
Clockwise Moments are: 5g x 1m ( weight of the shape is a force and its distance to A is half way between AD because its the centre of mass and because AD is 2m, then distance From A to 5g force is 1m)

Anticlockwise moments are: F x 2m ( that force at the top pulling it up) + F x 1m ( that force at the bottom pulling corner C to the right, if you look closely you can see that the force is already perrpedicular to A and distance to this force is AB )

M(A) 5g x 1 = Fx2 + Fx1

Hope It helped, Im self-studying m2
Right: the vertical distance from C to A is certainly 1m, but horizontally it's also 2m away. So, taking moments about A, don't you include that 2m distance aswell?

If not, then, for anticlockwise moments about A, any horizontal force applied along BC is simply (F X 1), right?
4. (Original post by fkhan100)
Right: the vertical distance from C to A is certainly 1m, but horizontally it's also 2m away. So, taking moments about A, don't you include that 2m distance aswell?

If not, then, for anticlockwise moments about A, any horizontal force applied along BC is simply (F X 1), right?
You need perpendicular distance from the force to the point you are taking moments about and I think the answer to your question is yes, AB (1m) x F is an anticlockwise moment I will attach a diagram to help you understand this question hopefully

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