# aqa M2

Watch
Announcements
#1
can anyone help with the following question?:

The diagram shows a uniform rod AB of weight W and length 2a. The rod rests with a on rough horizontal ground and leans against a rough fixed prism of semicircular cross-section of radius a. the coefficient of friction at both contacts is mu. When friction is limiting the rod makes an angle theta with the horizontal. show that

sin(theta)=sqrt((mu/(1+(mu squared)))

Question is from AQA M2 textbook, ex 2D, Q9, where there is also a diagram (and the final line is formatted better and easier to read!) if anyone has this book and it helps.
thanks
0
6 years ago
#2
...
Need the diagram. A scan/photo of the actual question, if you can, as well.

Then what have you done, and where are you stuck?
0
#3
i have no scaaning facilities at the minute, but have tried my best to copy the diagram from the textbook in paint (and the last line is in the box in the corner). When trying to work it out, i'd added normal reactions both where it meets the floor and where it meets the side of the prism (is supposed to be perfectly spherical, it's just badly drawn) and also frictional forces opposing what would otherwise be its direction of motion. I've tried taking moments about various points by resolving the forces, and equating friction for (mu x reaction). Have also tried using trig identities on it, but have been working on it for several days and not getting very far now.
0
6 years ago
#4
...
OK, makes sense now.

It's in limiting equilibrium, so if your normal reactions are R,S, then the frictional forces are.

So, standard techniques, resolve vertically, resolve horizontally, and take moments (in this case I did it about the lower end of the rod).

You will get 3 equations. And it's a process of eliminating what you don't want.

Spoiler:
Show

Eliminate W, one of the normal forces, and then you can cancel the other normal reaction, and re-arrange.

Post your initial equations if you wish, and we can check if your starting point is right.
0
#5
Is there any way of working out the distance between the centre of mass of the rod and the point of contact with the prism, as I'm struggling to take moments from the base of the rod without this?
0
6 years ago
#6
Is there any way of working out the distance between the centre of mass of the rod and the point of contact with the prism, as I'm struggling to take moments from the base of the rod without this?
Taking moments from the base of the rod doesn't require that distance - but can be worked out if you really need it by first doing:

The distance from the base of the rod to the point of contact with the prism.

If you consider the triangle formed by - base of rod, point of contact with prism, and centre of circle of prism, it's a right angled triangle. Side opposite theta has length "a" (a radius), Hence distance of base of rod to point of contact with prism is ....
0
#7
I'm thinking a/tan(theta), but I don't know if this would help me at all? I know there must be a more sensible alternative to this, but doesn't seem obvious.
0
6 years ago
#8
I'm thinking a/tan(theta), but I don't know if this would help me at all? I know there must be a more sensible alternative to this, but doesn't seem obvious.
Yes, or .

And in your moments equation the cosine part will cancel, as will the "a", leaving something nice and simple.
0
#9
I've got to the equation (i've put mu to represent mu here and x to represent theta)

(u^2)sinx + u(sin^2)x +(sin^2)x=sinx

by taking moments about B to get cosx=usinx, and then substituting it into equation formed by taking moments about the centre of the prism. Have also worked out that rod meets the prism halfway up the rod by taking moments about a (resolved forces rather than using horizontal distance).

however this doesn't seem to re-arrange to what the answer should be, so I'm wondering whether this is nearly right (and if so what I'm missing), or whether I've gone off on completely the wrong track?
0
6 years ago
#10
I've got to the equation (i've put mu to represent mu here and x to represent theta)

(u^2)sinx + u(sin^2)x +(sin^2)x=sinx

by taking moments about B to get cosx=usinx, and then substituting it into equation formed by taking moments about the centre of the prism. Have also worked out that rod meets the prism halfway up the rod by taking moments about a (resolved forces rather than using horizontal distance).
In bold above. This implies that the rod is in a fixed postion regardless of the angle of theta. So, you've made an error prior to this point.

I suggest posting your initial equations, and working. Can't tell where you've gone wrong otherwise.

I know it works out using moments about the base of the rod, resolving vertically, and resolving horizontally.
0
#11
I'd originally taken moments about A (this is obviously where I've gone wrong) but to do this I discounted both the frictional force where the rod meets the prism and the component of the weight parallel to the rod, as they act through A, and then setting aWcos(theta) equal to xWcos(theta) for the reaction, and therefore concluded that a is equal to x (i.e. that the reaction acts in the same line as the weight. I'm guessing this is where I've gone wrong-is the magnitude of the reaction at B equal to the component of weight perpendicular to the rod?
0
6 years ago
#12
I'd originally taken moments about A (this is obviously where I've gone wrong) but to do this I discounted both the frictional force where the rod meets the prism and the component of the weight parallel to the rod, as they act through A, and then setting aWcos(theta) equal to xWcos(theta) for the reaction, and therefore concluded that a is equal to x (i.e. that the reaction acts in the same line as the weight. I'm guessing this is where I've gone wrong-is the magnitude of the reaction at B equal to the component of weight perpendicular to the rod?
There are two components to the force at the point of contact with the prism, the frictional force parallel to the rod, and the perpendicular reaction, obviously perpendicular to the rod.

Spoiler:
Show

Where S is the perpendicular reaction to the rod at the point of contact with the prism

PS: A bit more maths, and a little less english. It's quite difficult to follow what you're doing otherwise.
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (288)
55.71%
I don't have everything I need (229)
44.29%