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

  • Thread starter Thread starter theneedtoknow
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
The problem involves determining how quickly two stacked boxes can travel 5 meters when pulled by a rope, considering static and kinetic friction coefficients. The solution focuses on the forces acting on the bottom box, calculating the net force and resulting acceleration without directly including the tension in the rope. The calculated acceleration leads to a time of 1.75 seconds for the boxes to cover the distance. The discussion highlights the surprising aspect that the tension does not need to be factored into the calculations, as the acceleration remains consistent across both boxes. Overall, the approach effectively simplifies the problem while yielding the correct answer.
theneedtoknow
Messages
169
Reaction score
0

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.
 
Physics news on Phys.org
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
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top