# Particle Kinematics with Friction

1. Feb 10, 2013

### aaronfue

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

A 1.5-lb brick is released from rest A and slides down the inclined roof. If the coefficient of friction between the roof and the brick is μ=0.3, determine the speed at which the brick strikes the gutter G.

3. The attempt at a solution

These are the equations that I got after setting up my FBD attached:

Fr(friction force) = 0.3*Normal force
ƩFx = 0.04658ax = -Fr + 1.5sin30°
ax = -11.81 $\frac{ft}{s^2}$

Then using the kinematic equation v2 = vo2 + 2a(s-so)

vo = 0
v2 = 2(11.81)(15)
v = 18.82$\frac{ft}{s}$

But that is not the correct velocity. I know I'm missing something. I just can't find it! The correct answer was 15.23 ft/s.

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Last edited: Feb 10, 2013
2. Feb 10, 2013

### tms

Your general approach seems sound, but it's hard to tell where you went wrong because you left so many steps out, and because you plugged in the numbers so early (when you do that you lose information about where the numbers came from).

I would suggest that you do it again using symbols instead of numbers. Pay particular attention to the frictional force.

3. Feb 10, 2013

### voko

First of all, you do not really care what the mass or weight of the brick is.

$$N = mg \cos \alpha \\ Fr = - \mu N = - \mu mg \cos \alpha \\ ma = mg \sin \alpha + Fr = mg \sin \alpha - \mu mg \cos \alpha$$

$m$ can now be eliminated from the equation.

4. Feb 10, 2013

### aaronfue

Okay. I have arrived at your final equation of:

a=gsinθ - μgcosθ (Eliminating the mass)

Now that I have the acceleration, I can plug this into my kinematic equation to obtain the final velocity as the brick completes the 15 ft or (s-so).

v2 = vo2 + 2*(gsinθ - μgcosθ)*(s-so)
vo2 = 0, since the brick started at rest.

If I were to plug in my values now, this would not be my answer.

Last edited: Feb 10, 2013
5. Feb 10, 2013

### CAF123

I get the required answer using that equation.
Alternatively, you can do this by energy methods to get:(Take zero of potential at start of gutter) $$mgh - F_fd = \frac{1}{2}mv^2\,\Rightarrow\,v^2 = 2(gh - \mu g\cos \theta d),$$ to get the same answer.

6. Feb 10, 2013

### aaronfue

You got 15.23 ft/s? I've been using two calculators (TI-89 & TI-83 plus) and have still not gotten that answer. I have made sure to be in the correct mode as well (degrees) but I'm not sure what my problem is. Would you be able to explain your calculations on your calculator or tell me what I'm doing wrong?

7. Feb 10, 2013

### CAF123

So you are in the correct mode. What value are you taking for $g?$. I converted everything into the more conventional units we use today. Would that be your problem?

8. Feb 10, 2013

### aaronfue

I'm using 32.2ft/s2 since we are using ft in the problem. I'll try to convert everything to m/s and 9.81m/s2.

9. Feb 10, 2013

### CAF123

That would also be valid. What number are you getting?

10. Feb 10, 2013

### aaronfue

For a second I was getting 11.20210. Now I am getting 21.799. I've attached a screenshot of my calculation.

Wait.......I've got it!!

I guess my order of operations was not going well. And I forgot to multiply μ*g.