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Simple Newtonian Mechanics question (infamous box and ramp question)

  • Thread starter mjdiaz89
  • Start date
  • #1
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Homework Statement


Ugh, rusty on my Newtonian Mechanics and need to refresh. I picked out this problem in my book:

Two boxes are on a 30 60 90 triangle. Box 1 is on the 60degree side and Box 2 on the 30degree side. The ramp has two different coefficient of static friction: u1 (for box 1) is 0.5 and u2 is 0.2. Both boxes are tied to a massless rope and a frictionless and massless pulley. Which way will the boxes slide?
Box 1 weighs 200lbs; Box 2 300 lbs.
Assume static initial conditions.


Homework Equations


F = ma
f= uN
{del}E = rho/epsilon zero (im kidding!)

The Attempt at a Solution


Ok, so I set up the 1st senario, the boxes slide with box 2 goign down first. Here's the work:
BOX 1)

E Fx = m1 a
T - fs1 - W1 sin(theta1) = m1 a

Box 2)
E Fx = -m2 a
T + fs2 - Wx2 = -m2 a

Combining the two equations (separating T and plugging it in the other):
[g * ( m1*cos(theta1)*u1 + m1*sin(theta1) + m2*cos(theta 2) -m2*sin(theta 2)]/[-(m1 + m2)] = a

a is -8.06 ft/s^2.


Checking my work, I do the complete opposite for the other scenario, where Box 1 goes down:
Box 1)
T + fs1 -W1*sin(theta 1) = -m1 a

Box 2)
T - W2*sin(theta 2) - fs2 = m2 a

Combining them and repeating, I get:

[g * ( m2sin(theta 2) + m2 g sin(theta 2)u2 + m1 cos(theta 1)u1 - m1sin(theta1))]/[-(m1+m2)] = a
a = 5.07 ft/s^2


Well...at least i got the signs to make sense from both accelerations...now where is my mistake as both magnitudes dont match? I cant find it :(




THANK YOU VERY MUCH!
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi mjdiaz89,

Homework Statement


Ugh, rusty on my Newtonian Mechanics and need to refresh. I picked out this problem in my book:

Two boxes are on a 30 60 90 triangle. Box 1 is on the 60degree side and Box 2 on the 30degree side. The ramp has two different coefficient of static friction: u1 (for box 1) is 0.5 and u2 is 0.2. Both boxes are tied to a massless rope and a frictionless and massless pulley. Which way will the boxes slide?
Box 1 weighs 200lbs; Box 2 300 lbs.
Assume static initial conditions.


Homework Equations


F = ma
f= uN
{del}E = rho/epsilon zero (im kidding!)

The Attempt at a Solution


Ok, so I set up the 1st senario, the boxes slide with box 2 goign down first. Here's the work:
BOX 1)

E Fx = m1 a
T - fs1 - W1 sin(theta1) = m1 a

Box 2)
E Fx = -m2 a
T + fs2 - Wx2 = -m2 a

Combining the two equations (separating T and plugging it in the other):
[g * ( m1*cos(theta1)*u1 + m1*sin(theta1) + m2*cos(theta 2) -m2*sin(theta 2)]/[-(m1 + m2)] = a

a is -8.06 ft/s^2.


Checking my work, I do the complete opposite for the other scenario, where Box 1 goes down:
Box 1)
T + fs1 -W1*sin(theta 1) = -m1 a

Box 2)
T - W2*sin(theta 2) - fs2 = m2 a

Combining them and repeating, I get:

[g * ( m2sin(theta 2) + m2 g sin(theta 2)u2 + m1 cos(theta 1)u1 - m1sin(theta1))]/[-(m1+m2)] = a
a = 5.07 ft/s^2


Well...at least i got the signs to make sense from both accelerations...now where is my mistake as both magnitudes dont match?
I have not checked all the numbers, but the magnitudes of your two cases should not match, because the frictional forces are entering the two cases differently.


What you have done in the first case is assumed that block 1 is moving upwards, and your result shows that if that is true then block 1 is accelerating downwards.

In the second case, you assumed that block 1 is moving downwards, and your result shows that if that is true then block 1 is accelerating downwards.


Remember that in general there is no reason for the acceleration and velocity to be in the same direction. In this problem they are, because it says the initial conditions are static. So in this problem your first case is not possible.


(It's the same idea as a single box sliding on a rough incline. The acceleration as it moves down the incline has a different magnitude than the acceleration as it moves up the incline.)
 
  • #3
11
0
Wow, good point. however, choosing a direction is only useful for makign sure your signs are consistent. The algebra says its going the other way, cant that be possible? Or is it so late that I cant see clear enough? o_O


hmm...i shall ponder your statements.
 
  • #4
alphysicist
Homework Helper
2,238
1
Wow, good point. however, choosing a direction is only useful for makign sure your signs are consistent. The algebra says its going the other way, cant that be possible? Or is it so late that I cant see clear enough? o_O


hmm...i shall ponder your statements.
I want to say again I haven't checked your numbers, since I have to leave soon.

But the thing to ask is what situation are you looking at in each of your cases? In case 1 you assume box 1 is sliding up the incline, so your results for case 1 only apply if that is true. Then you used the variable a in your equation in such a way that a positive result for a means that box 1 is accelerating up, and a negative result for a means that box 1 is accelerating down. Of course you found a negative result, which means that that result applies when box 1 is moving upwards and accelerating downwards. This is perfectly fine, if the boxes are given an initial shove to make that happen, but if they are released from rest this is not possible. And in this problem the boxes are initially static.

(I keep mentioning that I have not checked your numbers because I have to leave soon, but looking at your work quickly makes me think your result for case 2 may not be correct.)
 
  • #5
11
0
Yes there is one mistake i made when typing it: i did not factor out one of the g's. Typing mistake. Either way, thank you very much for you help.

Im kinda ashamed to ask this type of question as Ive taken physics for three years.... but i have what i believe to be a momentary brain fart... Mechanics was a while ago though... :(
 

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