Distance an object travels after sliding and falling from roof

[Natalie Johnson]

Homework Statement

[/B]
My physics teacher - "You may like this question, its very structured"
Calculate the distance a slate tile falls from a roof

The roof is angled at an angle 30 degrees.
mass of slate tile is 0.8kg
The length of the roof is L=6m
The height of the roof above the ground is h=10m
The object will slide down the roof a distance L before it like launched as a projectile through the air until it reaches the ground.
Coefficient of friction is assumed to be 0.5
g is for gravity
Find d, the distance it reaches from the roof is d.

Homework Equations

Friction force = Normal Contact force * coefficient of friction

Resolving in Y Direction
Normal Contact force = mass * g * cos (30)

Resolving in X Direction
ForceX= mass*g* sin(30) - Friction Force

Now find acceleration of object sliding down roof using F=m*a
a = ForceX / m

Now find the velocity at the point the tile leaves the roof using v^2 = u^2 + 2*a*s
a = ForceX/m
s = Length of Roof = 6m
v = ??

Now find the range distance the project will travel

d_max = sqrt ( (v^2 / g) + 2 * (v^2 / g) * h )

The Attempt at a Solution

Please can you comment if my above logic and equations is good. I can plug the numbers in no problem.

 

[PeroK]

It looks good apart from the last step, after it leaves the roof, which I don't follow.

Also, given what happens after it leaves the roof, perhaps you are better to keep x and y velocity components separate?

 

[Natalie Johnson]

It looks good apart from the last step, after it leaves the roof, which I don't follow.

Also, given what happens after it leaves the roof, perhaps you are better to keep x and y velocity components separate?

Hm yea I took an equation from my book for a range but I assume its wrong.
Do I need to work out time of flight before obtaining a distance?

 

[PeroK]

That would be the structured way to do it. Although you might think about the parabolic equation between y and x.

 

[Natalie Johnson]

That would be the structured way to do it. Although you might think about the parabolic equation between y and x.

Are you allowed to expand on that? I have been looking through my notes and unable to pin point what you mean

 

[PeroK]

Are you allowed to expand on that? I have been looking through my notes and unable to pin point what you mean

This is actually quite an interesting problem. If you stick with your original approach, you have a formula for the speed as it leaves the roof:

$v^2 = 2gLf(\theta)$ (Equation 1)

Where $f$ is some function of $\theta$ that I'll leave you to work out. Hint: it only involves $\theta$ and the coefficient of friction $\mu$. You have more or less done all this already in your original post.

Now, the slate leaves the roof at this point as a projectile with the above initial speed and at a downward angle of $\theta$ below the horizontal.

If you take the roof as $y = 0$ and downwards as positive $y$, then you get:

$y = (v\sin \theta)t + \frac12 gt^2$ and $x = (v\cos \theta)t$

If you combine these two, you should get an equation for $y$ as a parabolic function of $x$:

$y = Ax + Bx^2$ (Equation 2)

Where I'll leave you to work out $A, B$. Note: they are not constants but involve $\theta, g, v^2$. Note that getting rid of $t$ in this way can be a very useful technique.

Now, you can substitute equation 1 for $v^2$ into equation 2. That gets rid of $v^2$. But, do you notice anything else interesting?

Finally, if you set $y = h$, the height of the roof, you have a quadratic in $x$. You can plug in the values of $h$, $\theta$ and $\mu$ and solve this for $x$, the distance at which the slate hits the ground.

 

[Natalie Johnson]

I think I have got to the final part, Does the final equation mean the distance it travels on the ground is unrelated to the velocity at which it leaves the roof ?

 

[haruspex]

I think I have got to the final part, Does the final equation mean the distance it travels on the ground is unrelated to the velocity at which it leaves the roof ?

No, that cannot be true. What makes you think that? What answer do you get?
As a check, what should you get in the limit as L tends to infinity?

 

[PeroK]

I think I have got to the final part, Does the final equation mean the distance it travels on the ground is unrelated to the velocity at which it leaves the roof ?

The velocity as it leaves the roof is an intermediate variable. It's not an input to this problem. If you do a step-by-step approach you may have to calculate several intermediate variables. But, if you work through things algebraically, the intermediate variables should disappear.

It's up to you when you plug the numbers in. The full plug-and-chug approach might be:

Calculate all the forces; calculate the acceleration; calculate the speed/velocity as it leaves the roof; calculate the time of flight; calculate the horizontal distance where it hits the ground.

The full algebraic approach would be to generate an expression for $d$ that involves only the input parameters: $L, \theta, h, g$. Although, in this case, getting to a quadratic equation for $d$ is probably the place to stop and plug in the numbers.

I prefer keeping the algebra as long as possible for many reasons. It let's you see how the answer relates to the inputs. I find I can remember algebraic sequences: the same algebra turns up in very different problems. When the physics gets more advanced, you problems cannot be solved just by putting numbers into a calculator.

This might be a good problem to do both ways (and check you get the same numerical answer).