Forces and acceleration - two masses instead of one

[iJamJL]

Homework Statement

A red box and a blue box sit on a horizontal, frictionless surface. When horizontal force F is applied to the red box, its acceleration has magnitude a = 5.09 m/s2.

a) If force F is applied to the blue box, its acceleration has magnitude a = 1.24 m/s2. Find mred/mblue, the ratio of the mass of the red box to the blue box.

b) Now, the two boxes are glued together. If horizontal force F is now applied to the combination, find a-combo, the magnitude of the acceleration of the combination of boxes.

Homework Equations

F = ma

The Attempt at a Solution

I already correctly answered question a, which came to be a ratio of .2436. The second part, however, always confuses me because there has to be some sort of manipulation. Intuitively, when we put together the two boxes, we are combining their masses, making the total acceleration even smaller than what they were individually. This is true because we are applying the same amount of force to both of the objects. To get the numerical answer for the combined acceleration, I have no idea. I already know that adding the two accelerations together (two vectors in the same direction) and then multiplying it by the ratio (6.33 m/s^2 * .2436) gives me an incorrect answer, so that is probably a poor approach. I'm just stuck, like the rest of the problems I've had. lol it's quite frustrating.

 

[tiny-tim]

hi iJamJL!

A red box and a blue box sit on a horizontal, frictionless surface. When horizontal force F is applied to the red box, its acceleration has magnitude a = 5.09 m/s2.

a) If force F is applied to the blue box, its acceleration has magnitude a = 1.24 m/s2. Find mred/mblue, the ratio of the mass of the red box to the blue box.

b) Now, the two boxes are glued together. If horizontal force F is now applied to the combination, find a-combo, the magnitude of the acceleration of the combination of boxes.

…
The second part, however, always confuses me because there has to be some sort of manipulation. Intuitively, when we put together the two boxes, we are combining their masses, making the total acceleration even smaller than what they were individually.…

yes, that's right

so imagine there's a green box, whose mass equals the sum of the masses of the red and blue boxes …

what then? ​

 

[iJamJL]

hi iJamJL!


yes, that's right

so imagine there's a green box, whose mass equals the sum of the masses of the red and blue boxes …

what then? ​

Hi tiny-tim,

I have to admit, I'm still stuck at where I am. I'm imagining this scenario:

We have three hockey pucks that are colored blue, red, and green (to your suggestion ). The red one travels at 5.09 m/s^2 when I hit it with a certain force. When I hit the blue one, it travels at 1.24 m/s^2. If both of these pucks were to be hit at the exact same time, the blue one would be traveling .2436 times slower than the red one, leaving a gap that is increasing. Now if we introduce the third puck, and hit them all at the same time, the green one is even slower than the blue one because it is the heaviest.

I understand that the balance has to remain. If we have a certain force, that force will not change (in the formula F = ma). When mass increases, acceleration must decrease to keep F the same. I just don't know how to apply numbers to this scenario.

 

[tiny-tim]

hi iJamJL!

yes, everything you say is correct …

i don't understand why you're not getting this ​

I understand that the balance has to remain. If we have a certain force, that force will not change (in the formula F = ma). When mass increases, acceleration must decrease to keep F the same. I just don't know how to apply numbers to this scenario.

in other words: mass is inversely proportionate to acceleration

so if the masses were say 2 3 and 5,

the the accelerations would be in the ratios 1/2 1/3 1/5​

 

[iJamJL]

Sorry, I get mind blocks with these types of problems. I don't know why I don't understand it either

in other words: mass is inversely proportionate to acceleration

so if the masses were say 2 3 and 5,

then the accelerations would be in the ratios 1/2 1/3 1/5​

LOL I'm going to take two wild shots at what you're saying because my mind is struggling.

1) Since the mass is inversely proportionate to acceleration, I can say that the accelerations should then be 1/1.24 + 1/5.09, leaving me with a total of about 1.0 m/s^2?

2) The ratio is actually .2436, so that means it should be .2436/1.24 + .2436/5.09, leaving me with a total of .2443 m/s^2.

 

[tiny-tim]

1) Since the mass is inversely proportionate to acceleration, I can say that the accelerations should then be 1/1.24 + 1/5.09

no, the accelerations are 1.24 and 5.09

the masses should be in the ratios 1/1.24 and 1/5.09

 

[iJamJL]

I must be stupid, lol. Let me see if I can type this out.

If we start from the beginning, then we have the formula F = ma.

For red, F = m(red) * a(red)
F = m(red) * 5.09

Blue: F = m(blue) * a(blue)
F = m(blue) * 1.24

We combine them, calling it green:
F = m(green) * a(green)
F = [m(red)/1.24 + m(blue)/5.09] * a

lol I'm lost here. If we put the masses in ratios like that, then our formula F = ma becomes F = m(decrease)*a(increase). I don't see where the .2436 ratio applies.

Sorry, tiny-tim. I don't want to have you work it out for me, but I'm just not seeing the solution.

 

[tiny-tim]

For red, F = m(red) * a(red)
F = m(red) * 5.09

Blue: F = m(blue) * a(blue)
F = m(blue) * 1.24

So m(red) = F/5.09

m(blue) = F/1.24

 

[iJamJL]

So m(red) = F/5.09

m(blue) = F/1.24

:( I don't know what to do next..

 

[iJamJL]

I suppose if we add those two together, m(red) + m(blue) = F/5.09 + F/1.24, and then it'll be 5.1F/5.09, and then we divide it to make 1.0019=a

 

[iJamJL]

I think I might've come to another solution, but I do not know how to complete it. Let's take this try:

Red, F = m(red)*a(red)
Blue, F = [m(red)/.2436] * [a(red)*.2436]

If I add the masses, then m(red) + m(red)/.2436 = [1.2436*m(red)]/.2436 = 5.1149

Therefore, the acceleration is 1/5.1149 * 5.09 = .9951

 

[tiny-tim]

hi iJamJL!

(just got up :zzz:)

i don't follow your last post at all

this one is nearly correct (although difficult to follow at the end) …

I suppose if we add those two together, m(red) + m(blue) = F/5.09 + F/1.24, and then it'll be 5.1F/5.09, and then we divide it to make 1.0019=a

you'll find these questions easier if you write them out properly, one step at a time …

F = 5.09 m_a, so m_a = (1/5.09)F

F = 1.24 m_b, so m_b = (1/1.24)F

so m_a&b = m_a + m_b =(1/5.09 + 1/1.24)F = 1.003 F

so the acceleration of a&b = F/m_a&b = … ? ​