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Fullonrapist
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Physics
    #20702983 - 10/14/14 05:28 PM (9 years, 3 months ago)

a 100 kg box is pulled in the direction 20 degrees above the horizontal the box is moving in horizontal direction with an acceleration of 0.50 m/s2 and the coefficient of kinetic friction between the box and surface is 0.25. Find the force on the rope.

a 100 kg box is pulled by a rope parallel to an incline that is at an angle of 20 degrees with the horizontal. the box is moving with an acceleration of 0.50 m/s2 and the coefficient of kinetic friction between the box and the incline is 0.25. find the force on the rope

can you explain how to solve


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InvisibleDieCommie

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Re: Physics [Re: Pop-A-Cap]
    #20703316 - 10/14/14 06:46 PM (9 years, 3 months ago)

Newton's second law and free body diagrams.  Draw the free body diagram, appeal to newton's second law by summing the forces in each direction and equating to the appropriate mass and acceleration.


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Fullonrapist
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Re: Physics [Re: DieCommie]
    #20703425 - 10/14/14 07:08 PM (9 years, 3 months ago)

(100kg)(9.81m/s)(.25)=(100kg)(.50m/s)


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InvisibleDieCommie

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Re: Physics [Re: Pop-A-Cap]
    #20703437 - 10/14/14 07:12 PM (9 years, 3 months ago)

I don't know what that equality is, but its certainly wrong.  Punch each side into a calculator and see that the left side is not equal to the right.


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Fullonrapist
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Re: Physics [Re: DieCommie]
    #20703458 - 10/14/14 07:17 PM (9 years, 3 months ago)

But isn't the problem asking for the tension in the string? Fn=mg*Tsina


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InvisibleDieCommie

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Re: Physics [Re: Pop-A-Cap]
    #20703477 - 10/14/14 07:23 PM (9 years, 3 months ago)

You need to choose an appropriate coordinate frame, draw a free body diagram with all the forces, then sum the forces in the appropriate direction(s) and equate them to the appropriate acceleration(s).


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Fullonrapist
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Re: Physics [Re: DieCommie]
    #20703517 - 10/14/14 07:31 PM (9 years, 3 months ago)

For the first one the forces are friction, normal, weight and then the force pulling it. So (100kg)(9.81m/s)(0.25)=ma. Ma=(100kg)(0.50)=50


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InvisibleDieCommie

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Re: Physics [Re: Pop-A-Cap]
    #20703578 - 10/14/14 07:47 PM (9 years, 3 months ago)

Looks like gibberish to me... :shrug:  You dont even have the units right.

Did you draw a free body diagram?  How did you choose your coordinate system to be oriented?  What direction are you summing the forces in?  Why is there no trigonometry in your sum?  (In fact, why is there no sum at all?)

Your statement counts four forces, but I only see one term....  Did you collapse all the components into one term? But then where is the unknown force you are solving for?  Is m supposed to be equal to M?

I think you are trying to rush to the answer without thinking it through like you should.  You need to start with a free body diagram.


Edited by DieCommie (10/14/14 07:55 PM)


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Re: Physics [Re: DieCommie]
    #20703726 - 10/14/14 08:24 PM (9 years, 3 months ago)

Normal force is up, perpendicular to surface weight is down fk  is Toth left (retarding) and the string is being pulled at a 20 degree angle.I know the normal force is = to weight do they cancel


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Invisible4HO-DMT
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Re: Physics [Re: Pop-A-Cap]
    #20704255 - 10/14/14 10:23 PM (9 years, 3 months ago)

You're not listening to DieCommie. No the weight won't cancel the normal force, because of the rope pulling at a 20 degree angle. You should have two equations:

Sum(Fx)=m*ax

And:

Sum(Fy)=m*ay

Then you can solve for 2 unknowns with 2 equations. Make sure you are consistent with your coordinate system. Also, make sure your units are correct.


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InvisibleDieCommie

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Re: Physics [Re: 4HO-DMT]
    #20704279 - 10/14/14 10:29 PM (9 years, 3 months ago)

Depending on how you choose your coordinate system (and what is given), you may need two equations and you may only need one.  At this point I wouldn't worry about making the right tactical decision.  The practice could be useful by doing the problem once with a horizontal coordinate system relative to the ground and once with a 20deg coordinate system relative to the incline.


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Invisible4HO-DMT
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Re: Physics [Re: DieCommie]
    #20704530 - 10/14/14 11:27 PM (9 years, 3 months ago)

The first can be solved with one equation in the direction normal to the surface.

The second one though, I think needs two equations. Because of the incline, the normal force and the tension in the rope are unknowns. It is a simple set of equations, but formally, I think it requires two.


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Re: Physics [Re: 4HO-DMT]
    #20712526 - 10/16/14 06:46 PM (9 years, 3 months ago)

Remember, the normal force is essentially "zeroing" the sum of the forces for a direction an object is not accelerating in. My picture will be underneath, mostly to clarify the axis I'm using for this problem. First I drew a basic picture, mostly it's to help visualize the problem before I start a free body diagram. I then Drew the free body diagram, with my axis defined and noting the defined angles between the axis and forces.

Then I decomposed the vectors not laying along the axis I defined into the sum of two vectors that are parallel with this axis. I also went ahead and just wrote ukN=fk . That way I could draw a second free body diagram that contained only vectors parallel to these axis, this is an unnecessary step but it helps you avoid mistakes when you begin these problems. For me it always made it easier to sum the forces in given directions.

One could find T directly without finding N first, I merely did it so I could check my equation at the end to ensure I had the answer correct. The object is only Accelerating along x, and it is not accelerating along y. Thus the sum of y=0, and the sum of x=ma. The main reason I picked this coordinate system was to keep a parallel to an axis. That is generally how I treat problems like 2 as well, keeping acceleration along a single axis makes the F=ma portion of the summation easier, for me, because you can write the forces acting in opposite directions as being equal in magnitude without any additional constants effecting the equation.

Like I said, there is no reason to be required to find T first or second. Two variables, two equations. I like to write a step in all variables before I write it with only the two I'm focusing on though. It makes it easier for me to check for error. I hope this helped.



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