# M2 Moments Question

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#1
Struggling with some moments questions and here are some of them i would really love some help with
Attachment 635858For this one i'm really not sure how they took moments about O, particularly for mg. Quite confused about the bracket the mg is being multiplied by, i think it may be two distances added together from O, but the the image isnt clear in my head so if anyone could explain this perhaps with some diagram that would be great

for this one i honestly didn't even come up with a proper diagram, i'm really not following with the solution and would like some help here to.

Moments is probbaly the first chapter in m2 giving me this much trouble, does anyone know any good resources as well?
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4 years ago
#2
Struggling with some moments questions and here are some of them i would really love some help with
Attachment 635858For this one i'm really not sure how they took moments about O, particularly for mg. Quite confused about the bracket the mg is being multiplied by, i think it may be two distances added together from O, but the the image isnt clear in my head so if anyone could explain this perhaps with some diagram that would be great
First one:

In attached, line of action of mg force is in red, and perpendicular distance is the thick green line, which is the sum of the two parts, as you surmised.

1
4 years ago
#3
for this one i honestly didn't even come up with a proper diagram, i'm really not following with the solution and would like some help here to.
The given solution has a misprint; angle EBD is theta, not phi, in the diagram.

What help do you need? What don't you understand? The more specific you can be, the better.
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#4
(Original post by ghostwalker)
The given solution has a misprint; angle EBD is theta, not phi, in the diagram.

What help do you need? What don't you understand? The more specific you can be, the better.
1)How did they work out EF in triangle AEF
2) in the first step, when they state F/R=tanthetha? are they using a triangle of forces?

3) How did they work out the angle thetha besides R? and the dotted line beside it (which is Rcostheta) now is connected to a lower point which all the forces pass through could the question be solved doing moment about this point or any other method really?
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#5
(Original post by ghostwalker)
The given solution has a misprint; angle EBD is theta, not phi, in the diagram.

What help do you need? What don't you understand? The more specific you can be, the better.

I have another question here. They have asked to show that the horizontal force exerted by the ladder on the ground is independent of the mass of the brick "mg"
for this if resolve horizontally and show that S=F, then say F=uR then i find R=W+Mg, it shows that it is involved?
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4 years ago
#6
1)How did they work out EF in triangle AEF
EF = AF sin theta.

And AF = AC + CF = l + CF.
If you do some elementary geometry, you can show CF = l/2, and hence AF = 3l/2.

2) in the first step, when they state F/R=tanthetha? are they using a triangle of forces?
No, they don't say that. F/R = tan phi.

There are two different sets of angles referenced in this question, one is theta, and relates to the angle of inclination of the rod, and one is phi and relates to the forces.

The resultant of F and R, (F+R as vectors) must act along the line going through the common point (F). Hence F/R = tan phi.

3) How did they work out the angle thetha besides R? and the dotted line beside it (which is Rcostheta) now is connected to a lower point which all the forces pass through
It's phi, not theta. And I can't make sense of that.

Almost certainly, but I'm not working it through to find out.
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4 years ago
#7

I have another question here. They have asked to show that the horizontal force exerted by the ladder on the ground is independent of the mass of the brick "mg"
for this if resolve horizontally and show that S=F, then say F=uR then i find R=W+Mg, it shows that it is involved?
It's involved in the vertical reaction, but that doesn't tell you anything about the horizontal/frictional force. And note F=uR, is only true in limiting equilibrium.
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#8
(Original post by ghostwalker)
EF = AF sin theta.

And AF = AC + CF = l + CF.
If you do some elementary geometry, you can show CF = l/2, and hence AF = 3l/2.

No, they don't say that. F/R = tan phi.

There are two different sets of angles referenced in this question, one is theta, and relates to the angle of inclination of the rod, and one is phi and relates to the forces.

The resultant of F and R, (F+R as vectors) must act along the line going through the common point (F). Hence F/R = tan phi.

It's phi, not theta. And I can't make sense of that.

Almost certainly, but I'm not working it through to find out.
Do you mean how they worked out phi doesn't make sense or my question?
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4 years ago
#9
Do you mean how they worked out phi doesn't make sense or my question?
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#10
(Original post by ghostwalker)
how did they geometrically work out the angle i circled?
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4 years ago
#11
how did they geometrically work out the angle i circled?
Combining F and R reduces the problem to three forces, the tension in the string, the weight and the reaction at the ring.

The lines of action of the three forces in equilibrium must go through a common point.

The tension in the string and the weight are enough to define that point.

Hence the line of action of the reaction at the ring must also go through that point, that is the dotted red line on your diagram, from which the angle phi arises.
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#12
(Original post by ghostwalker)
It's involved in the vertical reaction, but that doesn't tell you anything about the horizontal/frictional force. And note F=uR, is only true in limiting equilibrium.
The question says "in order to enable the ladder to rest in equillbrium in the position described above, a brick is attached at the bottom of the ladder without touching the ground"
It says in equilibrium, so why is equating S=F, then say F=uR then i find R=W+Mg wrong? I mean at least it should be because it says that it's independent of the mass of the brick
or can i not say F=uR because they didnt say limiting, so instead it'd be F<=u(W+Mg)
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4 years ago
#13
No problem - it's important to stick with a problem until you understand.

The question says "in order to enable the ladder to rest in equillbrium in the position described above, a brick is attached at the bottom of the ladder without touching the ground"
It says in equilibrium, so why is equating S=F, then say F=uR then i find R=W+Mg wrong? I mean at least it should be because it says that it's independent of the mass of the brick
or can i not say F=uR because they didnt say limiting, so instead it'd be F<=u(W+Mg)
Resolving horizontally S=F, that's fine.

The question says show the horizontal force exerted by the ladder is independent of the mass of the brick, i.e. the force F.

R, as you've worked out, does depend on the mass of the brick.

You're correct in that you cannot say F= uR, but rather
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