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Acceleration of a Block at an Incline

  • Thread starter sess
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  • #1
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Homework Statement


A rope applies a force T = 500N as shown as the 30kg block moves 14m along the incline. The length of the incline is unlimited. The coefficient of friction between the block and the plane is 0.33
What is the acceleration of the block?

Homework Equations


[itex]\Sigma[/itex]F = ma


The Attempt at a Solution


So for this I tried -
Fgx=mgsinθ - Fgy=mgcosθ
ƩFx=max
Tsinθ-mgsinθ-μkFN=max
So, after that I try to plug in FN=mgcosθ-Tcosθ
And then solve, then I take that answer as the force and divide it by the mass, 30kg, which would give me the answer.

However, I know the answer to this question and the one that I keep getting is wrong. Any help you guys could provide would be really appreciated.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
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No worries - the trick is to align your axis to the slope so the block accelerates in the +x direction. That way you only need to resolve gravity into components - not the tension.

I'm guessing the rope pulls the block up the incline?
Have you been given the slope of the incline or do you have to figure that out?
When showing equations, careful with hyphens: they are used as subtraction signs.

tension - gravity - friction = ma.

So why do you have Tsinθ in yours?
 
  • #3
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Oh, yes, sorry, I posted this question when I was going to bed last night and I was very tired. :P The slope of the incline would be 28°. I see what you mean about not resolving the tension into components now. And yes, the rope would be pulling it up the incline.

So how would I go about solving the FN? Would that just be mgcosθ?

So it would be:

T-mg-μkFN=ma
500N-294N-(0.33)(260)=120N
120N/30kg = 4m/s2

But that's not the right answer, so the only thing that I can think that I'm doing wrong would be the FN.
 
  • #4
Doc Al
Mentor
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So how would I go about solving the FN? Would that just be mgcosθ?
Yes.
T-mg-μkFN=ma
You want force components parallel to the incline. What's the component of the weight parallel to the incline?
 
  • #5
3
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Ahhh, I see what you are saying:

T-mgsinθ-(μk)(mgcosθ)=ma

So that would be-

500N-138N-85.8N=ma
276.2N/30kg
9.2m/s2

Which is the correct answer. Thanks guys, I really appreciate your help.
 
  • #6
Simon Bridge
Science Advisor
Homework Helper
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Well done.

You were having trouble figuring out when to resolve components and when not to.
To get this right every time, draw a picture of what is happening, and draw arrow on it for the forces. Then draw your x-y axes on the diagram so that most of the forces line up with them (usually you want to put one axis is the direction you'd expect the movement to happen.) Remember that you can tilt axis however you like - so you line them up so the math is easiest ;)

Now the rule is - any remaining vectors that do not line up with any axis get resolved into components.

In the example you showed us - the friction and the tension lined up with the x axis, the normal-force lined up with the y axis, leaving only gravity pointing at an angle to the axes. Thus, only gravity need be resolved to components.

If you draw the picture like I described, you'll see how it all fits together right away.
 

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