How Fast Can Two Blocks Move Without Sliding Apart?

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Homework Help Overview

The problem involves two blocks with given coefficients of static and kinetic friction, and it asks for the minimum time required for both blocks to move a distance of 7.00 m without the top block sliding off the lower block. The masses of the blocks and the forces acting on them are provided.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations of maximum static and kinetic friction forces, and the net force acting on the blocks. There is a focus on understanding the implications of these forces on the acceleration of the bottom block and the overall motion.

Discussion Status

Some participants have provided calculations and insights into the forces involved, while others are seeking clarification on how to properly apply these forces to the bottom block's motion. The conversation indicates a productive exploration of the problem, though consensus on the next steps has not been reached.

Contextual Notes

Participants are navigating the implications of using the mass of only the bottom block in subsequent calculations, as well as the relationship between the forces acting on both blocks.

chris097
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Homework Statement

The coefficient of static friction is 0.636 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.138. Force F causes both blocks to cross a distance of 7.00 m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.14 kg and the mass of the upper block is 2.80 kg?

http://capa.physics.mcmaster.ca/figures/kn/Graph08/kn-pic0832.png


The attempt at a solution

Fmax= UsN
=(.636)(2.80*9.81)
=17.4696 N

F(bottom) = UkN
= (.138)((2.80 + 1.14) * 9.81)
=5.3339

Fnet = 17.4696 - 5.3339
=12.1357

F=ma
a=F/m
=12.1357/(1.14+2.8)
=3.0801

d=v1*t + .5at^2
t=square root of (d/(.5a))
= square root of (7/(.5(3.0801)))
=2.13 s


where am i going wrong?
thank you
 
Last edited by a moderator:
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chris097 said:
Fmax= UsN
=(.636)(2.80*9.81)
=17.4696 N

F(bottom) = UkN
= (.138)((2.80 + 1.14) * 9.81)
=5.3339

Fnet = 17.4696 - 5.3339
=12.1357

this is the maximum net force that can accelerate the bottom block.
you should consider only the acceleration of the bottom block
in your next calculation.
 
Thank you.

But what do you mean consider on the bottom block. Do you mean use only the weight of the bottom block?
 
chris097 said:
Thank you.

But what do you mean consider on the bottom block. Do you mean use only the weight of the bottom block?

yes. Since you found the maximum net force on the bottom block, then you need to use the mass and the acceleration of the bottom block as well in F = ma
 

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