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Im ready for this test except I cant get started on these 2 questions.

  • Thread starter davie08
  • Start date
  • #1
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Homework Statement



A force of 20N,180degrees (W) and 30N,-110degrees (S20degrees (W) ) are both acting on a 3.0kg mass at rest. If there is 10N of friction:calculate the acceleration.

Homework Equations



I think you just need kinematic equations so here they are:
V2=V1+a*t
d=V1t+1/2a*t^2

The Attempt at a Solution


Im going to try the R theta and change that to x and y I just wanted to put this up first stop me if this is wrong.
 

Answers and Replies

  • #2
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wait you need forces equations:
Fn=m*a
Fn=Fa+Ff
Fg=m*g
Ff=m*F perpendicular sign
 
  • #3
Doc Al
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A force of 20N,180degrees (W) and 30N,-110degrees (S20degrees (W) ) are both acting on a 3.0kg mass at rest.
What's the total applied force?
If there is 10N of friction:
Which way does the friction act?
calculate the acceleration.
First find the net force acting on the mass.
 
  • #4
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well the friction would be negative right since its going against the object
 
  • #5
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oh wait the total applied force are the first numbers
 
  • #6
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so I use Fn=Fa+Ff how do I change the first numbers into something I can use
 
  • #7
Doc Al
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how do I change the first numbers into something I can use
To add the two applied forces, first break them into their components. Then you can add the components and find the new resultant.
 
  • #8
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im not sure how you chnge them dont you just use pol on the calculator and then F to get them but what about the S20degrees (W)
 
  • #9
minger
Science Advisor
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This problem I think is a little easier than you're making it. You started on the correct approach. Break each force into it's components. Add the two forces together to get the resultant components. From these forces, calculate the resultant force vector.

You now have a force mangitude and direction. Focus for a moment on the magnitude. You're given the friction force. This force acts opposite of motion, so you don't need to worry about direction. Simply subtract the friction force from the total resultant force.

You now have a resultant force and direction; divide by mass to get acceleration. I get 10.45 m/s² as the magnitude of the acceleration. Remember acceleration is a vector, you'll need to specify the direction as well.
 
  • #10
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im not sure how you chnge them dont you just use pol on the calculator and then F to get them but what about the S20degrees (W)
You shouldn't really need a calculator to break a vector into its components. Just trigonometry. Well, I guess you'll need the calculator to solve for the sin and cos of the angles...

Breaking up a vector into its components should be second nature by this point in the semester/school year.
 
  • #11
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using that pol F stuff makes it 10 times easier
 
  • #12
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im just going to hope this isnt on the test lol I actually understand most of this unit now so I should be ok on the test
 
  • #13
minger
Science Advisor
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First of all, you need to take heed of what Jack said. Breaking vectors into components needs to be second nature by now. You can't always rely on calculator to add things in polar coordinates, hell you can't always use a calculator.
 

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