Free body diagram, inclined plane, and finding V from x vs t graph

AI Thread Summary
A free body diagram of a block on a 30-degree inclined plane must accurately represent the forces acting on it, including friction, normal force, and gravitational force. The friction force should be equal to the component of gravitational force acting down the plane since the block is stationary, meaning it balances out the forces. To find velocity from a position vs. time graph, one should calculate the slope of the graph at the desired time, which represents the derivative dx/dt. If the graph is linear around that point, using a small interval to calculate the average slope can yield the instantaneous velocity. Understanding these concepts is crucial for accurately solving the physics problems presented.
jcpwn2004
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Homework Statement


Draw and label the free body diagram of a block on a plane inclined 30 degrees from the horizontal. There is friction and the block is stationary. The magnitues of the vectors must be correctly scaled and the directions correct.

And then my other question is how to find velocity from a positon vs time graph at a given time.


Homework Equations





The Attempt at a Solution



http://img201.imageshack.us/my.php?image=inclinedc5.jpg

Here's my free body diagram. I guess i just want to see what the magnitudes should be. Should Fk be just slightly bigger than mgsinO since it is stationary? Also N should be equal to Fperpindicular right? And how big should mg be?

For the second part I don't understand how to find the velocity at a given point. Like if I'm supposed to find the velocity at say 2 seconds, would i just try to find the slope between 1.9 and 2.1 or what?
 
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Your drawing doesn't have arrows, and vectors should have direction of course.

As to your position:time graph, think for a moment what velocity is.

Isn't it Δx/Δt ?

And what happens when you make that Δ very small such as you might to find the Velocity at a particular point in time?

What does that Δx/Δt become as Δ -> 0 ?
 
LowlyPion said:
Your drawing doesn't have arrows, and vectors should have direction of course.

As to your position:time graph, think for a moment what velocity is.

Isn't it Δx/Δt ?

And what happens when you make that Δ very small such as you might to find the Velocity at a particular point in time?

What does that Δx/Δt become as Δ -> 0 ?

So for my drawing if i drew arrows would it be accurate?

And then for the second part what would i do because if Δ -> 0 then it will be undefined. SHould i do like a limit or something?
 
jcpwn2004 said:
So for my drawing if i drew arrows would it be accurate?

And then for the second part what would i do because if Δ -> 0 then it will be undefined. SHould i do like a limit or something?

Judging by your relative magnitudes I suppose you're OK then.

What I was getting at with the Δt thing was that you should recognize that as the derivative (tangent) to the position function at the time in question. Specifically the Δx/Δt as Δt -> 0 is dx/dt = Velocity at any point on the graph.
 
LowlyPion said:
Judging by your relative magnitudes I suppose you're OK then.

What I was getting at with the Δt thing was that you should recognize that as the derivative (tangent) to the position function at the time in question. Specifically the Δx/Δt as Δt -> 0 is dx/dt = Velocity at any point on the graph.

So if my problem said to find the velocity from this graph at 2 seconds I would find dx/dt at 2? I understand the velocity at 2 is 20 m/s i just don't know how to show my work for it. Would i do like V = Δx/Δt so V = (50-30)/(2.5-1.5) = 20 or something like that?

Also what could i do to improve the magnitudes for the free body diagram?
 
With no magnitudes all you can do is show the variables.

And yes if you know the slope throughout a range about t=2 is a straight line, just choose some easily calculated interval to yield a result.
 
LowlyPion said:
With no magnitudes all you can do is show the variables.

And yes if you know the slope throughout a range about t=2 is a straight line, just choose some easily calculated interval to yield a result.

Don't I know the magnitudes since it's stationary though? Shouldn't the friction force be larger than mgsin0 since it isn't sliding down the ramp? Also don't I know that the normal force is equal to the perpindicular force (mgcos0)?
 
jcpwn2004 said:
Don't I know the magnitudes since it's stationary though? Shouldn't the friction force be larger than mgsin0 since it isn't sliding down the ramp? Also don't I know that the normal force is equal to the perpindicular force (mgcos0)?

That's right. Relatively speaking. But keep in mind that Friction is really only an Fmax. And in your case is only supplying a force equal to the downward along the plane force of m*g.
 
LowlyPion said:
That's right. Relatively speaking. But keep in mind that Friction is really only an Fmax. And in your case is only supplying a force equal to the downward along the plane force of m*g.

thanks for all the help :)
 
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