Propulsive force and force at interface

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Homework Help Overview

The problem involves two blocks of different masses placed on a horizontal table, with a constant horizontal force applied to the larger mass. The scenario includes frictional forces acting on one block while the other experiences no friction, and both blocks are accelerating at a specified rate.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law (F=ma) to analyze the forces acting on the blocks. Initial attempts to set up equations and identify forces are noted, along with suggestions to visualize the problem through diagrams.

Discussion Status

Participants have engaged in calculations regarding the net force and have proposed values for the propulsive force and the force at the interface between the blocks. There is a recognition of the need to consider the forces acting on both blocks and the implications of friction.

Contextual Notes

Participants are navigating the complexities of the problem, including the effects of friction and the relationship between the forces acting on the two blocks. There is an emphasis on ensuring that the applied forces are sufficient to achieve the given acceleration.

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Homework Statement


Two blocks, one of mass 6 kg and the other of mass 2 kg, are placed in contact on a horizontal table, and a constant horizontal force is applied to the 6 kg mass. There is a constant frictional force of 18 N between the 6 kg mass and the table, but no frictional force between the 2 kg mass and the table. the blocks accelerate at 3 m/s2.
a. what is the propulsive force applied to the 6 kg mass?
b. what is the force at the interface between the two blocks?

Homework Equations


F=ma

The Attempt at a Solution


I honestly have no idea where to start I have read the book and looked at my notes and I am not sure where to start.
Fnet=ma
fnet= F-18[/B]
 
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Start by drawing a large diagram on which you mark everything relevant. Examine it closely, looking for inspiration.

Post it here.
 
upload_2014-10-16_3-11-25.png
 
Now you can see what is going on!

Apply your Fnet = ma formula to the situation.
 
Fnet=mneta
F-18=mneta
F-18=(6+2)(3)
F-18=24
F=42N
 
is this correct?
 
That looks right.

Next, what force must be being applied to the 2kg body to cause it to be accelerating at this rate? (Remember, if you don't apply sufficient force to it it won't accelerate at this rate, and if you were to apply more than the required force it will accelerate faster.)
 
Will that force be the force at the interface. Since 2 kg mass experiences no friction
Fnet=ma
Fnet=2(3)
F=6N
 
Will that force be the force at the interface. Since 2 kg mass experiences no friction
Fnet=ma
Fnet=2(3)
F=6N
Yes that force can only come from the bigger mass pushing on that leading body.
 
  • #10
So a) what is the propulsive force applied to the 6kg mass? 42N
and b) what is the force at the interface between the two blocks? 6N
 
  • #11
It's always a good idea to associate a direction with the description of vector quantities (even when the direction may not be explicitly sought).

So, for each answer decide whether the horzontal force is directed to the right or the left.
 

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