Finding the minimum force F which must be exerted

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Homework Help Overview

The problem involves a block with mass 6 kg on an inclined plane of a wedge with mass 18 kg, where the friction coefficient is 0.27. The goal is to determine the minimum force required to prevent the 6 kg block from sliding down the incline. The scenario includes considerations of forces acting on both the block and the wedge, as well as the effects of gravity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on both the wedge and the block, suggesting the use of free body diagrams (FBDs) to clarify the relationships between forces. There is uncertainty about how to incorporate the effects of the 6 kg block into the calculations for the force on the wedge.

Discussion Status

Some participants have provided guidance on using free body diagrams to analyze the forces involved, while others express confusion about the setup and calculations. There is no explicit consensus on the correct approach, and multiple interpretations of the problem are being explored.

Contextual Notes

Participants note discrepancies in the free body diagrams presented, particularly regarding the normal forces acting on the blocks. There is an emphasis on the need to accurately represent these forces to progress in solving the problem.

rando25
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Homework Statement


Given: The friction between the block with mass 6 kg and the wedge with mass 18 kg is 0.27 . The surface between the wedge with mass 18 kg and the horizontal plane is smooth (without friction). The acceleration of gravity is 9.8 m/s^ 2. A block is released on the inclined plane (top side of the wedge). What is the minimum force F which must be exerted on the 18 kg block in order that the 6 kg block does not move down the plane?
29zen9h.jpg


Homework Equations


Fnet = ma
and possibly(?): Ff = uFn

The Attempt at a Solution


Having the wedge in the problem is confusing me.
The forces acting on the x-axis of the 18 kg block are F applied and Fg parallel
The forces acting on the x-axis of the 6kg block are F frictional and Fg parallel.
Fnet = ma / Fnet = 0
Fa + Fg(parallel) = 0
I think I am solving for Fa (18kg), I'm not sure.
Fa = Fgsin20
Fa = (18kg)(9.8m/s2)sin20
Fa = 60.33 N
I know the 6kg block plays a role, but i don't know how or where to incorporate it.
I'm really confused, any help is appreciated.
 
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Having the wedge in the problem is designed to get you to think.

Start with FBDs for both the wedge and the 6 kg block and write equations of motion for each AFTER you draw the FBDs.
 
OldEngr63 said:
Having the wedge in the problem is designed to get you to think.

Start with FBDs for both the wedge and the 6 kg block and write equations of motion for each AFTER you draw the FBDs.
 
OldEngr63 said:
Having the wedge in the problem is designed to get you to think.

Start with FBDs for both the wedge and the 6 kg block and write equations of motion for each AFTER you draw the FBDs.
upload_2014-11-28_18-22-38.png

this is my work so far. I know I am looking of F at this point, but I am still not sure where to go from here? (unless my FBDs are incorrect...)
 
Your FBDs do not agree with each other. For example, the Fn you show on the 6 kg block is upward and to the left; the Fn you show on the wedge is straight up. There should be one Fn that is the opposite of the other. You also have not distinguished the fact that there are two normal forces involved. And it goes on...

I suggest that you try again on the FBDs.
 

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