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# Mechanics Question Help watch

1. Could someone help me out with the second part of this question?
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2. (Original post by Firefox23)
Could someone help me out with the second part of this question?
you need to attach it...
3. What ave you tried so far?
4. (Original post by Zacken)
What ave you tried so far?
I initially tried to find the acceleration using F=man with F as the weight down the plane - the friction up the plane and used s=ut +1/2at^2 with s as 5a and u as 0
5. draw a clear diagram first then label all of the forces along each plane and you should be able to do this rather easily.
6. (Original post by Firefox23)
I initially tried to find the acceleration using F=man with F as the weight down the plane - the friction up the plane and used s=ut +1/2at^2 with s as 5a and u as 0
Okay, so we're taking down the slop as positive then - this means we have

so we need only find acceleration then.

Let's see - we resolve down the plane to get where we can get R easily, simply resolve perpendicular to the plane to get that

So our force down the plane is where .

Hence you get (after dividing by m) .

Using the suvat equation yields

That's the time down the slope, they are asking for the total time - so you need to add the time to go up the slope as well.

That's a basic suvat solve for let's call it , which you should get as
Spoiler:
Show

Then you add and pull out a common factor.
7. (Original post by Zacken)
Okay, so we're taking down the slop as positive then - this means we have

so we need only find acceleration then.

Let's see - we resolve down the plane to get where we can get R easily, simply resolve perpendicular to the plane to get that

So our force down the plane is where .

Hence you get (after dividing by m) .

Using the suvat equation yields

That's the time down the slope, they are asking for the total time - so you need to add the time to go up the slope as well.

That's a basic suvat solve for let's call it , which you should get as
Spoiler:
Show

Then you add and pull out a common factor.
Thanks so much I forgot to add the time up the slope to get the total time. Any chance you could help me with this question? I'm not sure where to start.
Attached Images

8. (Original post by Firefox23)
Any chance you could help me with this question? I'm not sure where to start.
Start at A.

Since you have symmetry, what's the vertical reaction at A?

As A and E are as far apart as possible what's the frictional force going to be at A?

What's the vertical component in AB?

You now have sufficient information to work out the angle of AB, and hence the horizontal displacment of B.

Now go on to consider B and BC.
PS Not worked that bit through.
9. (Original post by ghostwalker)
Start at A.

Since you have symmetry, what's the vertical reaction at A?

As A and E are as far apart as possible what's the frictional force going to be at A?

What's the vertical component in AB?

You now have sufficient information to work out the angle of AB, and hence the horizontal displacment of B.

Now go on to consider B and BC.
PS Not worked that bit through.
The tension in each section of the string will be different right? Also what do you mean by the vertical component in AB? This is what I tried . . .
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10. (Original post by Firefox23)
The tension in each section of the string will be different right?
The tension in the upper strings will be different to that in the lower strings.

Also what do you mean by the vertical component in AB? This is what I tried . . .
The tension in AB can be split into vertical and horizontal components.

You can write down the vertical reaction at A straight off. What's the total weight? Then by symmetry vertical reaction at A is ...?

Then as per my previous questions.

Note: There is no force, only friction and the horizontal component in AB.
11. (Original post by ghostwalker)
The tension in the upper strings will be different to that in the lower strings.

The tension in AB can be split into vertical and horizontal components.

You can write down the vertical reaction at A straight off. What's the total weight? Then by symmetry vertical reaction at A is ...?

Then as per my previous questions.

Note: There is no force, only friction and the horizontal component in AB.
This is what I've tried but I still seem to have too many unknowns . . .
12. (Original post by Firefox23)

This is what I've tried
Doesn't seem that way from your working - unless you've done something since my last post, but haven't posted it.
13. (Original post by ghostwalker)
Doesn't seem that way from your working - unless you've done something since my last post, but haven't posted it.
So my working's correct?
14. (Original post by Firefox23)
So my working's correct?
Sorry, didn't see your new working at the top of the post - will have a look now.
15. (Original post by Firefox23)
So my working's correct?
You have ignored what I suggested.

The total weight of the mass is 8mg.
The vertical reactions at A and E must add to 8mg.

By symmetry the vertical reaction at A will be half that, i.e. 4mg.
16. (Original post by ghostwalker)
You have ignored what I suggested.

The total weight of the mass is 8mg.
The vertical reactions at A and E must add to 8mg.

By symmetry the vertical reaction at A will be half that, i.e. 4mg.
Does the tension in AB not affect the vertical reaction? Sorry I didn't misread the question and didn't notice that the A and E had different masses to B,C and D
17. (Original post by Firefox23)
Does the tension in AB not affect the vertical reaction? Sorry I didn't misread the question and didn't notice that the A and E had different masses to B,C and D
The tension in AB does effect the vertical reaction. The vertical reaction at A, 4mg, arises as a result of the weight of A, mg, and the vertical component of the tension in AB.
18. (Original post by ghostwalker)
The tension in AB does effect the vertical reaction. The vertical reaction at A, 4mg, arises as a result of the weight of A, mg, and the vertical component of the tension in AB.
I think I understand R=4mg and it also = mg +tcostheta
19. (Original post by Firefox23)
I think I understand R=4mg and it also = mg +tcostheta
Yep.

In the first case you look at the structure as a whole, and in the second you just focus on the details at A.
20. (Original post by ghostwalker)
Yep.

In the first case you look at the structure as a whole, and in the second you just focus on the details at A.
Thanks I've worked it through now, any chance you could help me out with this question if you have time? I've tried to do velocity triangles and I've tried i and j notation but I can't seem to get the right answer. Its question 14.

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