# Homework Help: Equilibrium of a uniform bar inclined at an angle

1. Jul 22, 2016

### 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.

2. Jul 22, 2016

### Simon Bridge

It doesn't. The angles are associated with different objects.
They have, when they say the plane and thr peg are "smooth". This means zero friction.
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.

3. Jul 23, 2016

### Taniaz

What do you mean by different objects?

4. Jul 23, 2016

### Nidum

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

5. Jul 23, 2016

### 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?

6. Jul 23, 2016

### Taniaz

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

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7. Jul 23, 2016

### 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 ?

8. Jul 23, 2016

### Nidum

wait a few minutes .

9. Jul 23, 2016

### Simon Bridge

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 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?

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.

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

10. Jul 23, 2016

### Nidum

11. Jul 23, 2016

### 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 ?

12. Jul 23, 2016

### Taniaz

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

13. Jul 23, 2016

### 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.

14. Jul 23, 2016

### Nidum

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

15. Jul 23, 2016

### 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.

16. Jul 23, 2016

### Nidum

Last edited: Jul 23, 2016
17. Jul 23, 2016

### Taniaz

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

18. Jul 23, 2016

### Nidum

Bar is represented by line AB .

Plane is represented by line DE .

Last edited: Jul 23, 2016
19. Jul 23, 2016

### 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

20. Jul 23, 2016

### 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?

21. Jul 23, 2016

### Nidum

Yes - that's all you need .

22. Jul 23, 2016

### Taniaz

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

23. Jul 23, 2016

### Nidum

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

24. Jul 23, 2016

### 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)?

25. Jul 23, 2016

### 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.