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Moments Q M2 watch

1. I added the red arrows in the above diagram. They are supposed to represent the normal reactions at A, C and D but they are guesses and I am not sure if there actually are normal reactions at these points?

Also, do frictional forces act at A, C or D? And why/why not?

Thanks
2. (Original post by sabre2th1)
...
There are certainly normal components to the reactions at those points, but the reactions are not normal.

And no, there are no frictional forces, nothing is sliding over anything else: At each point the two meeting components are attached, in one form or another.

The forces in the string will be along the line of the string.
3. (Original post by ghostwalker)
There are certainly normal components to the reactions at those points, but the reactions are not normal.

.

What do you mean with the above?

Thanks
4. (Original post by sabre2th1)
What do you mean with the above?

Thanks
E.g. At D the force is acting along the line of the string. So the reaction will be opposite to that. It can be resolved into a horizontal component and a vertical component.

Haesights is also correct regarding A.
5. (Original post by Haesights)
Its hinged so the reaction force has a vertical and a horizontal component (at A)

So at A, are the vertical and horizontal components equal?
6. (Original post by ghostwalker)
E.g. At D the force is acting along the line of the string. So the reaction will be opposite to that. It can be resolved into a horizontal component and a vertical component.

Haesights is also correct regarding A.

Are the arrows for the reaction force at D correct?
7. (Original post by Haesights)
There is no reaction force at D, only A.
If there was no reaction at D there could be no tension in the string.

(Original post by sabre2th1)
Are the arrows for the reaction force at D correct?
No, those are the components for the tension in the string. The reaction force's components would be in the opposite direction; up and to the left.

And at A the horizontal and vertical components will, in general, be different.

Also at C, the force acting will be along the line of the string.

If you're only interested in the beam, then you don't need the forces at D. What's the actual question you're trying to answer?
8. (Original post by Haesights)
I thought the tension force would be acting up and to the left?
Tension acts towards the centre of the structure under tension.

So at the bottom end of the string the tension would be up and to the left, and at the top end of the string it would be down and to the right. And the reaction at the top end, which is opposite to the latter, would be up and to the left.
9. (Original post by Haesights)
O like how they act in opposite directions when the car is towing something?

Mechanics wording is so confusing
Yes, in a towbar, under tension, the forces in the towbar are towards the centre of the towbar. It's the same situation in that regard.

Mechanics wording is exacting, like most maths questions.

Although my use of the word centre here isn't exactly right, it's good enough for most questions.
10. (Original post by ghostwalker)
No, those are the components for the tension in the string. The reaction force's components would be in the opposite direction; up and to the left.
So when annotating this diagram, would I have to include BOTH the components for the tension AND the reaction force's components on my diagram? (I mean if I didn't then wouldn't my calculations get muddled up?)

I checked Examsolutions earlier but he didn't draw neither the tension nor the contact forces at D

Also at C, the force acting will be along the line of the string
The force ''along the line of the string'' will be the tension right? So the vertical red arrow I drew at C is still correct?

If you're only interested in the beam, then you don't need the forces at D. What's the actual question you're trying to answer?
I have to show that Tension = mg(root13)

Thanks a lot !
11. (Original post by sabre2th1)
So when annotating this diagram, would I have to include BOTH the components for the tension AND the reaction force's components on my diagram? (I mean if I didn't then wouldn't my calculations get muddled up?)
You're likely to get more muddled if you have both the forces and the reactions in your diagram, particularly if you put them in component form.

I checked Examsolutions earlier but he didn't draw neither the tension nor the contact forces at D
If they are not necessary for the question, then there is no necessity to include them.

The force ''along the line of the string'' will be the tension right?
That's correct.

So the vertical red arrow I drew at C is still correct?
Not by itself. That's just the vertical component of the tension; you'd need the horizontal component as well. Or alternatively, and simpler, just put in the tension along the string.

I have to show that Tension = mg(root13)
Since you're only going to be working with the beam, you only need to show the forces acting on the beam. As such, you don't need all the reactions, nor do you need the forces at D. Which is why the given solution has left them out.

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