Acceleration of 2 boxes sliding down ramp - please

AI Thread Summary
The discussion revolves around determining the acceleration of two boxes sliding down a ramp. Participants emphasize the importance of drawing free body diagrams (FBDs) for each box to analyze the forces acting on them, including gravitational, normal, and frictional forces. It is clarified that both boxes share the same acceleration since they move together, and Newton's second law is applied to derive the equations of motion. The acceleration of the system is calculated to be 1.8 m/s², leading to a force of 18 N acting on package B. The conversation highlights the interconnectedness of the boxes' motions and the necessity of considering their combined effects to solve the problem.
nukeman
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Homework Statement



Please refer to the diagram. I have having trouble with B: "Determine the acceleration of the system. Explain your method, and identify and laws you are applying"

Ok so for A:, I drew two FBD's. For package A there is Fg, Fnormal, Ffriction, and F from box B - Correct?

Box B - Fg, Fnormal, Ffriction, and FfromboxA - Correct?

So now B is where I am having trouble. I am not sure how to find the acceleration of the SYSTEM ?


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Homework Equations





The Attempt at a Solution

 
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nukeman said:
Ok so for A:, I drew two FBD's. For package A there is Fg, Fnormal, Ffriction, and F from box B - Correct?

Box B - Fg, Fnormal, Ffriction, and FfromboxA - Correct?
OK. (Hopefully you drew the directions correctly.)

So now B is where I am having trouble. I am not sure how to find the acceleration of the SYSTEM ?
Newton's 2nd law. (Analyze force components parallel and perpendicular to the incline.)
 
Well, box A in front has a higher coefficient of friction, and so it will always be moving slower than box B. So the motion of box B is affected by the motion of box A, and vice versa (i.e. box B is pushing on box A because box A won't get out of its way).

So the motion of the system is really determined by the motion of box A or B, since they move together. When you draw the FBD of box A, first assume that box B isn't there, and include all relevant forces, then add in the force box B exerts on A, and that should describe the motion of the system.
 
pergradus said:
Well, box A in front has a higher coefficient of friction, and so it will always be moving slower than box B. So the motion of box B is affected by the motion of box A, and vice versa (i.e. box B is pushing on box A because box A won't get out of its way).

So the motion of the system is really determined by the motion of box A or B, since they move together. When you draw the FBD of box A, first assume that box B isn't there, and include all relevant forces, then add in the force box B exerts on A, and that should describe the motion of the system.

Ok I see what you are saying.

I am just having trouble figuring out the acceleration. What method do I use to figure this out?
 
The acceleration of either box A or B is the acceleration of the system - they are both moving with the same acceleration.
 
pergradus said:
The acceleration of either box A or B is the acceleration of the system - they are both moving with the same acceleration.

Yes I get that... since they are moving together, they have the same acceleration.

BUT - how do I figure out the actual acceleration?
 
nukeman said:
Yes I get that... since they are moving together, they have the same acceleration.

BUT - how do I figure out the actual acceleration?

You need to write your equations of motion for each box. After doing this what do you have in the inclined direction? Remember Newton's third law...
 
kjohnson said:
You need to write your equations of motion for each box. After doing this what do you have in the inclined direction? Remember Newton's third law...

so my first step would be to choose a cordinate system, and then delevope an equation?
 
nukeman said:
so my first step would be to choose a cordinate system, and then delevope an equation?

Yep. This is the case for tackling almost any problem. So what do you get?
 
  • #10
kjohnson said:
Yep. This is the case for tackling almost any problem. So what do you get?

Give me a sec... We just started doing these (FBD's to equations) and IM not 100% comfortable yet doing them.

Ill post it in a sec here.
 
  • #11
Ok drew it out, with coordinate system, and a table. Not sure if this is correct. Can yuou look over my table?

t6v15e.jpg
 
  • #12
Draw them out as separate and draw Free body diagrams then equations for each block. Your table looks ok for the entire system but you should label 'Fgravity' as 'Fparallel' or something like that indicating that it acts parallel to the ramp. The value of 'Fgravity' is just 'mg' and it acts vertical. On the other hand 'Fparallel' and 'Fperpendicuarl' are the components of 'Fgravity' in the x and y directions.
 
  • #13
By the way you can solve for the acceleration of the blocks by solving it as a system vs writing equations for each block. But then you will still have to write and equation for one of the blocks to find the force 'F' that acts between them...
 
  • #14
So, if I go the system route, is my table correct?

To add to my table the ma of x would be what? and what would the ma of y be?
 
  • #15
Can anyone confirm my table is correct?

what is the ma of x and the ma of y

Still don't know how to find those :(

Thanks so much guys!
 
  • #16
How are you able to find the frictional force on the system?
 
  • #17
SammyS said:
How are you able to find the frictional force on the system?

Its just the sum of the friction forces for both blocks. I was not 100% sure how the system approach would work but it seems that if you do it as a system it is the same as if you wrote an equation for each block and added them together canceling out the interaction force 'F' between them.
 
  • #18
Now for C

Now package A and package B are sliding down a ramp together, so they have the exact same acceleration. The acceleration of the system is 1.8 m/s^2 and the mass of package B is 10kg

So, is it simply F = ma

F = (10kg)(1.8) = 18 N ?
 
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