Equilibrium of a uniform bar inclined at an angle

[Taniaz]

1. A uniform bar AB of length 3 m and weight 40 N rests in equilibrium with the lower end A on a smooth plane inclined at an angle alpha to the horizontal, where tan alpha = 3/4 and a point C of the rod against a smooth peg. If the rod makes an angle theta with the horizontal, where tan theta = 7/24, find (a) the magnitudes of the reactions at A and C, (b) the distance AC.2. Nc (sin 37) + Na = w so Nc (3/5) + Na = 40
Nc (cos 37) = F so Nc (4/5) =F
40 (cos 37)(1.5) = Nc (x) where x is AC so 40 (4/5)(1.5) = Nc (x) so 48 = Nc(x) 3. The first thing I didn't understand was why has the question provided 2 angles at which the plane is inclined to?
Secondly, they haven't provided any coefficient of friction or any other variable which would have made solving these equations easier. You need at least one more variable to get any of the required quantities.

 

[Simon Bridge]

The first thing I didn't understand was why has the question provided 2 angles at which the plane is inclined to?

It doesn't. The angles are associated with different objects.

Secondly, they haven't provided any coefficient of friction or any other variable which would have made solving these equations easier.

They have, when they say the plane and thr peg are "smooth". This means zero friction.

You need at least one more variable to get any of the required quantities.

Id prefer to know which direction the angles are measured in and how big the peg is... but you probably have standard assumptions for those.
I think you need to go over the description more carefully and sketch a diagram.

 

[Taniaz]

What do you mean by different objects?

 

[Nidum]

The plane is sloping at one angle and the rod is sloping at a different angle .

 

[Taniaz]

I don't understand their orientation with respect to one another. The rod is supporting the peg? So it'll be two angles a certain distance apart but in the same direction?

 

[Taniaz]

So this is the diagram I now have. I think we'll be using theta for the angle at C

 

[Nidum]

It is a very poorly worded question but there are some clues to help us interpret it .

Which is inclined at the greater angle - the rod or the plane ?

 

[Nidum]

wait a few minutes .

 

[Simon Bridge]

I don't understand their orientation with respect to one another.

It is not entirely clear to me either... but you can make some headway by understanding that it is the plane that is at $\theta$ to the horizontal, and the rod that is at angle $\alpha$ to the horizontal.

The rod is supporting the peg?

The problem statement says that the rod is resting on the peg and the plane - maybe these are the rod's support: that is part of what your calculations will find out. But consider, if there were no plane, where would point C have to be?

So it'll be two angles a certain distance apart but in the same direction?

Maybe both angles are measured anticlockwise from the horizontal? That would be conventional - but the problem statement does not say. An alternative would be that $\theta$ is clockwise from the horizontal while $\alpha$ is anticlockwise. Part of the problem will be that you have to figure out which is intended.

Additionally:
If $\tan\alpha = 3/4$ then you can make a 3-4-5 triangle.

 

[Nidum]

 

[Nidum]

1 and 2 both seem to be valid interpretations of the question geometry but realistically 1 is the only one which can allow stability ?

 

[Taniaz]

Nidum, where is the rod in your diagram? What do the plus signs stand for?

 

[Taniaz]

Simon could you please take a look at my diagram? Do we resolve any forces on the rod? We don't know anything about it other than the fact that it's supporting the peg.
I'm getting these equations once again: Nc (x) = w cos (alpha) (1.5 m) and Nc sin(alpha) + Na = w where Nc, Na and x are unknown.
Really don't know where we're going to use theta - the angle between the rod and the horizontal.

 

[Nidum]

The +'s are just frame delimiters from the CAD .

 

[Taniaz]

I'm sorry but I'm still not sure what's going on. All I'm getting are these equations?
Nc (x) = w cos (alpha) (1.5 m) and Nc sin(alpha) + Na = w where Nc, Na and x are unknown.
Really don't know where we're going to use theta - the angle between the rod and the horizontal.

 

[Nidum]

 

[Taniaz]

Where is the rod in your drawings?? And where are the angles? That is what I don't get!

 

[Nidum]

Bar is represented by line AB .

Plane is represented by line DE .

 

[Nidum]

Bar is at angle tan^-1 BF/AF relative to horizontal . Given as tan^-1 7/24

Plane is at angle tan^-1 EG/DG relative to horizontal . Given as tan^-1 3/4

 

[Taniaz]

Ok that makes a lot of sense now. Thank you.
So our components are now Na (normal at A), the weight of the bar, and Nc (normal at c)? Do we take any other components due to the plane into account?

 

[Nidum]

So our components are now Na (normal at A), the weight of the bar, and Nc (normal at c)?

Yes - that's all you need .

 

[Taniaz]

But there's one problem, 3 unknowns and not enough equations.
And of what use is the angle alpha angle to us?

 

[Nidum]

The direction of the line of action of the normal force at A depends on the slope angle of the plane .

 

[Taniaz]

Ohh right ofcourse, because the plane is inclined so the normal acts perpendicular to it. So the vertical component of Na is going to be Na cos (alpha)?

 

[Taniaz]

So the equations are:
1) w = Na (cos alpha) + Nc (cos theta) -----> 40 = (4/5) Na + (24/25) Nc from Σƒ y = 0
and
2) Nc (x) = w cos (theta) (1.5) ------> Nc(x) = 48 where x is the distance AC from Γa=0
But it still can't be solved.

 

[Taniaz]

Am I missing something with these equations?

 

[Nidum]

Draw a free body diagram of the bar and let us see it .

 

[Taniaz]

This is the diagram that I have with the components. Anything missing?

 

[Taniaz]

So am I missing something?

 

[Nidum]

Forces in X direction
Forces in Y direction
Moments .

 

[Taniaz]

I've got the sum of forces in y and the moments. Just don't have the sum of forces in x.

Are the ones I have correct?

 

[Nidum]

I'll let you work this problem out the most obvious way for now .

There is a more elegant method possible - perhaps we'll look at that later if you are interested .

 

[Taniaz]

Yes I'd like to know what the way method is as well?

 

[Nidum]

Use the diagram in post #16 as a basis for doing a clear free body diagram for the bar .

Show us the three equations that you should now have and , if possible , your final solution .

 

[Taniaz]

I already drew the free body diagram on the basis of post 16 and attached it in one of my previous posts and I already mentioned the 2 equations I get from the forces in the y direction and the moment around A, just doing it for the x direction now and that should give me the answer.