Pulling mass on flat plane at an angle.

AI Thread Summary
The discussion revolves around a physics problem involving a block being pulled on a rough horizontal surface at an angle. For part A, the normal force is derived by considering the vertical components of the forces, specifically accounting for the applied force's vertical component. In part B, the coefficient of friction is expressed in relation to the normal force and gravitational force. Participants express confusion about graphing speed and displacement over time, noting that constant acceleration can be used to derive these relationships. Additionally, a question about the maximum acceleration before losing contact with the surface is raised, emphasizing the importance of understanding the forces involved.
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Homework Statement


A block of mass m is pulled along a rough horizontal surface by a constant applied force of magnitude F that acts at an angle theta to the horizontal. The acceleration is a. Express all algebraic answers in terms of m, F, theta, and a.
A. Derive an equation for the normal force
B. Derive an expression for the coefficient of friction between the block and surface.
Sketch the graphs of speed v and displacement x as teh functions of time if the block started at rest at x=0 and t=0


Homework Equations


Fy=(Applied force)(Sin(theta))
Fx=(Applied force)(Cos(theta))
Fnet=(mu)(Normal force)


The Attempt at a Solution



Well for part A, I figure that since it was being pulled at an angle that would provide for a lower normal force.

Normal Force=(Fg-Fy) *Fy is the y component of the force vector.
For the coefficient of friction, I don't know if I am supposed to derive an equation that soves for mu, or just the equation that factors in mu.

Here is what I have for part B:
Ff=u[Fg-Fy]

And for the graphs I am kind of lost. Thanks for any help!
 
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=(Fg-Fy) *Fy is

Why you multiplied it by Fy?
It's kind of very rare(in dynamics) to multiply forces . . .

Ff=u[Fg-Fy]
here u is the coefficient of friction so, you do have it factored out.

Add all the horizontal forces(or analyze them), you will find an equivalent for Ff as net 'a' is provided.

For graphing, either use work-energy relationship or
simply use kinematicss,
you know 'a', and it is constant
 
the *Fy was sort of a side info comment. It isn't part of the equation. Are you saying that my friction equation is correct? I'm sorry but I'm still somewhat confused on the graphs as well.
 
as Ff = uFn
and so you know Fn is W-Fy
so yea, it is correct.

so acc. = a
and so how its related to x,v,t?
You don't need forces or dynamics for this question.

Ignore that Work-energy, I thought they are asking for v,x relationship

But, You cannot draw graph, as you are not given any numerical value.
 
Well...
A=v/t

This is an old AP question that asks for a graph. I would agree too but there is some type of solution out there.

Another question that is on the back that I didn't notice was:

-If the applied force is large enough, the block will lose contact with the surface. Derive an expression for the magnitude of the greatest acceleration Amax that the bllock can have and still maintain contact with the ground.
 
So, Fy = Fg

and F = Fy.sintheta


you may draw an approximation graph, like if it's linear draw a straight line...
 
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