# How long does it take mass m to fall distance h?

1. Sep 16, 2010

### vandersmissen

1. The problem statement, all variables and given/known data

[PLAIN]http://rawrspace.com/Capture.JPG [Broken]

So when you draw out a force diagram, we find these two equations
mg-T=ma
rewriting in terms of T
T=mg-ma
and
T-MgSin$$\vartheta$$-f=Ma
rewriting in terms of T
T=Ma+MgSin$$\vartheta$$+f

I set the two equations equal to each other so I could solve for the acceleration since it is as rest when it starts (I figured that means that initial velocity is 0 )
setting them equal I get:
mg-ma=Ma+MgSin$$\vartheta$$+f
Then I moved them so that I could start factoring out terms:
g(m-MSin$$\vartheta$$)-f=a(M+m)
finally, solving for a:
[g(m-MSin$$\vartheta$$)-f]/(M+m)=a

I also know that y-y_0=V_0t-(1/2)at^2
Since y final is 0 and y_0 is h, I get this equation:
-h=-(1/2)at^2
So this means that
(2h/a)^(1/2)=t

Pluging in a that I solved for above, I get

{(2h)/[g(m-MSin$$\vartheta$$)-f]/(M+m)}^(1/2)=t

That was my answer of the time it takes for the mass m to fall a distance h. Does that seem correct to you, or am I overlooking something ?

Last edited by a moderator: May 4, 2017
2. Sep 16, 2010

### kuruman

It looks right to me except your answer has f in it and you are specifically required to write the answer in terms of m, M, h, θ and μ. So how can you get rid of f and introduce μ?

3. Sep 16, 2010

### vandersmissen

Thank you for taking a look. I don't know why I wrote it like that on here, on my paper I have μN for the frictional force instead of just the f for frictional force..

Thank's again for taking a look.

4. Sep 16, 2010

### kuruman

You're still not there because μN won't do either. What is N in terms of the given quantities m, M, h, θ?

5. Sep 16, 2010

### vandersmissen

Well N is the normal force so it is the opposite component the gravitational force which is Mg so it would be -Mgcos$$\vartheta$$ wouldn't it (assuming I take my g to be positive)? That would still leave me with a g though which isn't in those terms. Is it okay to have g since it is a constant ? If so then I would replace -f with μMgcos$$\vartheta$$ (it would be positive correct, as -f = -μN = -μ(-Mgcos$$\vartheta$$) )

So I would like to write my answer as

{(2h)/[g(m-Msin$$\vartheta$$)+μMgcos$$\vartheta$$]/(M+m)}^(1/2)=t

6. Sep 16, 2010

### kuruman

Don't forget that, in your original expression, f stands for the magnitude of the frictional force. That is a positive number. So f = μN = Mgcosθ. You already took care of the direction when you wrote down T-Mgsinθ-f=Ma.

7. Sep 16, 2010

### vandersmissen

So It would be μMgcosθ correct since N = Mgcosθ , and f = μN

So that means that it would be -μMgcosθ in the problem instead of -f

t={[(2h)(M+m)]/[g(m-Msinθ)-μMgcosθ]}^1/2

I believe that would be correct as it is now in the correct terms, I moved the M+m to the numerator. Does that seem correct now, If it was -f which took care of the magnitude since it subtracted in the original force equation. Then replacing it with the information I derived above should be correct.

8. Sep 16, 2010

### kuruman

Right.

9. Sep 16, 2010

### vandersmissen

Thank you for helping me understand where I went wrong.