What Is the Minimum Static Friction Coefficient to Keep a Block Motionless?

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Homework Help Overview

The problem involves a 4.0 kg block on a table connected to a hanging 1.0 kg mass, focusing on the minimum static friction coefficient required to keep the block motionless. The discussion includes concepts of static and kinetic friction, force diagrams, and the balance of forces in a statics context.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of force diagrams and the direction of forces based on chosen coordinate systems. Questions arise regarding the use of kinetic friction in a static scenario and the balance of forces acting on the block.

Discussion Status

Participants are exploring the transition from a dynamics problem to a statics problem, with some guidance provided on the nature of static friction and the balance of forces. There is an acknowledgment of the need to consider limiting friction in the context of the applied forces.

Contextual Notes

There is confusion regarding the application of kinetic friction in the static scenario, and participants are questioning the assumptions about force directions and the balance of forces acting on the block.

Raiden9
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Hello, I deeply need help on this problem. ANY help would be appreciated.

Suppose a hanging 1.0 kb lab mass is attached to a 4.0 kg block on the table.
A) If the coefficient of kinetic friction, uk is 0.20, what is the accereration of the block?
B) What would be the minimun value of the coefficient of static friction, in order for the block to remain motionless?

I have my force diagrams made, equations also. What I am getting confused on is do I add the remaning unbalanced forces, Ff and Fg or is the Ff negative? Any help please!

Thank You!
 
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The forces involved can be positive or negative depending on how you choose your coordinate system. For example, if you choose +y to be down, then mg is positive because it acts in the same direction as +y. The key to friction force is that it always acts in a direction opposite to motion. So, if you chose +x to be the direction the block on the table moves, then friction will be in the -x direction. The key is to be consistent with the coordinate system you decide to use.
 
Oh I see. I have my axis as a traditional axis. So, I add my unbalanced forces according to the axis positive or negative values? I'm still stuck on the part b problem though, so any tad bit of help would help.
 
Part b) is a statics problem. All the forces balance.

Have you got all the forces on the 4kg block ?
 
Yes, I have the Fn y+, Fgy-, Ftx+, and Ffx-...
 
Would I use the coefficient of kinetic friction from the last problem for the Ff in part b? I am really confused! Someone help please!
 
It's now a statics problem! So the coefft of kinetic friction doesn't get involved. There's no movement.

What you have is limiting friction.
The block is static. When a force is applied to a static block, then there will always be static friction equal and opposite to the applied force on the block.
As the force increases, the static friction increases until the limiting condition is reached. The static friction reaches a maximum value and the block is just about ready to move. The ratio of this friction/applied force to the normal reaction is called the coefft of static friction.

This is the state that you have reached with the block. There is an applied force on the block. There is a static friction force opposing this. There is a normal reaction. That's all you need.
 
I got it! Thanks!
 

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