Finding the minimum force F which must be exerted

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The discussion revolves around calculating the minimum force required to prevent a 6 kg block from sliding down an inclined plane on a wedge. The wedge has a mass of 18 kg and a friction coefficient of 0.27, while the surface beneath it is frictionless. Participants emphasize the importance of drawing free body diagrams (FBDs) for both the wedge and the block to accurately analyze the forces at play. Confusion arises regarding the correct representation of normal forces in the FBDs, which are crucial for solving the problem. Overall, the thread highlights the need for clarity in force analysis to find the solution effectively.
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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