How Quickly Can Two Stacked Boxes Travel 5 Meters When Pulled by a Rope?

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Homework Help Overview

The problem involves two stacked boxes, with a focus on determining the time it takes for them to travel a distance of 5 meters when pulled by a rope. The scenario includes considerations of static and kinetic friction coefficients affecting the motion of the boxes.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve the problem by analyzing the forces acting on the bottom box, leading to a calculated time of 1.75 seconds without considering the tension in the rope. Some participants question whether the entire system's mass should be considered, while others suggest that focusing on one box is sufficient for determining acceleration.

Discussion Status

The discussion is active, with participants exploring different interpretations of the forces involved. There is acknowledgment of the calculated time being consistent with posted solutions, but questions remain regarding the role of tension and the treatment of the system as a whole.

Contextual Notes

Participants are navigating assumptions about the forces acting on both boxes and the implications of static versus kinetic friction in the context of the problem. There is a noted surprise regarding the lack of tension consideration in the original poster's approach.

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Homework Statement


a 4kg box rests on top of a 3kg box. There is a horizontalrope attached to the top box that can pull the boxes. What is the shortest amount of time the boxes can travel 5 meters together if the static friction coefficient between them is 0.6 and the kinetic friction coefficient between the bottom box and the ground is 0.2?


Homework Equations





The Attempt at a Solution



Well I am pretty sure I've already solved the problem, it's just that my solution kind of surprises me so I want to ask a few questions.
So I solved it by considering the bottom box
It has a force in the direction of movement due to static friction, so that force would be 4g x 0.6 (4g is the magnitude of the force of the top box on the bottom one due to gravity)
it also has a force opposin movemet that would be -7g x 0.2 (-7g is the force of both boxes on the ground and therefore the normal force of the groudn on the lower box is 7g)

So acceleration of b is Fnet / mass of b = (0.6x4g - 0.2x7g )/3 = g/3

so shortest amount of time for 5 meters is 5 = 0.5 x g/3 x t^2 -----> t = root (5 / (0.5g/3)) = 1.75 seconds


Now my question is ... I'm kind of surprised that my answer doesn't involve considerin the forces actin on the top box at all...such as the tenion of the rope...When i first looked at the problem i thought i would have to create a system of linear equations (from the x and y components of the forces acting on the 2 boxes) with my unknowns being the tension of the rope and the acceleration and I would have to solve for both. However...I didn't end up considering any forces from the top box and only needed its mass...Did i do something wrong? Cause my answer seems legit to me.
 
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since the blocks are moving together, the acceleration is the same for each block...so it's only necessary to solve for acceleration in one of the blocks, in this case the 3kg block since we don't know the tension force
 
Don't you need to accelerate both boxes?

The total mass being accelerated is 7 kg and not 3 kg then isn't it?
 
That would be if you consider the entire system together... if you are looking at only the x-component of forces acting on the lower box by itself, you would only consider the mass of the lower box...i just checked the posted solutions and the answer is indeed 1.75s ... i guess i just find it suprising that the tension didn't come into play...i thought i would have to find a maximum possible tension before the boxes start slipping, and find the acceleration that corresponds to that max tension
 

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