# Homework Help: Kinda tricky inclined plane problem

1. Oct 17, 2016

### Jrlinton

1. The problem statement, all variables and given/known data
Mass 1 of 1 kg is resting on an inclined plane of 30 degrees. A string is attaached to the mass and ran through a pulley at the top of the plane and attached to a suspended mass 2 of 2 kg. How far must the inclined plane be rotated until mass 1 starts to move down the incline.
2. Relevant equations

3. The attempt at a solution
So I am trying to use all of my knowledge of inclined planes to no avail. Firstly the force of friction must be the forced imposed upon mass1 by the gravitational force of mass 2 minus the gravitational force in the direction of the incline (m1*g*sintheta+Ffriction*g2*g)

2. Oct 18, 2016

### haruspex

There is no mention of friction. Even if you allow friction I do not see how this will lead to a solution.
Is there anything about the statement of the problem that strikes you as paradoxical? (I would suggest making sure you have quoted it exactly.)
The paradox can be resolved, but it involves a little lateral thinking.

3. Oct 18, 2016

### CWatters

Perhaps consider what happens if the slope was increased to say 90 degees/vertical.

4. Oct 18, 2016

### haruspex

Which way?

5. Oct 18, 2016

### Jrlinton

Okay if the angle was I increased to 90 degrees then the mass of the second block would pull the first up, so the angle has to be between 30 and 90 degrees as the sin of theta increases in regard to mass 1.

6. Oct 18, 2016

### haruspex

Why does there have to a solution anywhere in that range? What would happen at 30 degrees, with no friction?

7. Oct 18, 2016

### CWatters

The sin of any angle is less than one - so how does that increase the effect of mass 1 above that of the vertical case?

8. Oct 18, 2016

### Jrlinton

Either this question has no solution, I am remembering wrong, or I am approaching it wrong. If there was a given friction coefficient, then the sum of the gravitational force (mgsinΘ) and the frictional force (umgcosΘ) of mass1 would be equal to the force of Mass 2 (mg).
Simplifying I get
sinΘ+ucosΘ=2
Assuming the coefficient of friction is reasonable, this is impossible.

9. Oct 18, 2016

### haruspex

Since we are told nothing of friction, assume it is irrelevant. Consider my hint (that is what it was) in post #4.

10. Oct 19, 2016

### Jrlinton

Okay so lets suppose that mass1 iss 2kg and mass2 is 1 kg and the coefficient of static friction is 0.5.
Would you not subtract the force of the friction (0.5*2*9.81*sinΘ) from the gravitational force on Mass1 (2*9.81*cosΘ) and set that equal to the weight of mass 2 (9.81*1)?
(19.62cosΘ)-(9.81sinΘ)=9.81?

11. Oct 19, 2016

### haruspex

This is the reverse of the OP statement, right? Are you now supposing that was wrong?
As I have hinted, there is a solution to the question as originally given, but it's a bit of a trick.
I think it will be helpful to keep it general. Let's say μ.
Friction acts to oppose relative motion of surfaces in contact. If you assume that m1 would otherwise slide down, then yes, giving (19.62cosΘ)-(9.81*2μsinΘ) <=9.81, or more simply, 2cosΘ - 2μsinΘ <= 1. If we further assume that friction barely holds it, then = 1.
Note that if we assume no friction there is a sensible solution to that equation.

12. Oct 20, 2016

### Jrlinton

What I know is that mass 1 of 2kg is sitting on an inclined plane. There is a string attached to the mass, ran through a pulley and attached to a suspended mass of 1 kg. The coefficient of static friction is 0.5. At what angle will mass 1 begin to move down the slope?
I did as I previously described and set 2cosΘ-sinΘ=1. Firstly I must be forgetting some trig identity as I am unaware as to have to solve for Θ, second when plugging the given answer choices, the equation does not prove to be true.

13. Oct 20, 2016

### haruspex

Ok, this is at last making sense. You originally had the two masses swapped, and omitted to mention friction was given.

There is a trick you can use here. Are you familiar with equations like sin α cos β - cos α sin β = ....? Can you see how to coerce the left hand side of your equation into that form?

14. Oct 20, 2016

### Jrlinton

This is wrong. Somehow I messed up bad.
It should be 2sinΘ-cosΘ=1

15. Oct 20, 2016

### haruspex

OK, but my reply above still applies.

16. Oct 20, 2016

### Jrlinton

Honestly I'm unaware as to how that identity can be used, could be the lack of sleep or something. Anyway I rearranged the equation to 2tan(theta)=1+1/cos(theta) and got 53.1 degrees.

17. Oct 21, 2016

### haruspex

What is the obvious reason it does not immediately fit into that form?

18. Oct 21, 2016

### Jrlinton

Limiting theta to the first quadrant

19. Oct 21, 2016

### haruspex

No. You have an equation 2sinΘ-cosΘ=1. What simple fact prevents you from writing the left hand side in the form sin α cos β - cos α sin β?

20. Oct 21, 2016

### Jrlinton

the multiplier of 2 in regards to sinΘ

21. Oct 21, 2016

### haruspex

Right, it is out range for cos β. More exactly, the sum of squares of the coefficients is 22+12=5, whereas cos2β+sin2β = 1. So what can you do to both sides of your equation to get around that?

22. Oct 21, 2016

### vela

Staff Emeritus

You could rewrite this version as
$$\sin\theta = \frac{1+\cos\theta}{2}.$$
There's an obvious trig identity that comes into play here.

23. Oct 21, 2016

### Jrlinton

That's effectively what I did earlier and further altered the equation to tan(theta)=0.5(cos(theta)+1)

24. Oct 21, 2016

### haruspex

I can see where Vela is trying to lead you, and it does work for this question. But there is a more generally useful trick. Please try to respond to post #21.