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Math Help - Force problem.

  1. #1
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    Force problem.

    Force questions always confuse me because of the N. (newtons?)

    Two forces, 1 and 2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 55.5 N and F2 = 42.5 N. What is the horizontal acceleration (magnitude and direction) of the block?

    please help break this down for me.

    http://img171.imageshack.us/img171/3181/0411vf7.gif
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    Force questions always confuse me because of the N. (newtons?)

    Two forces, 1 and 2, act on the 5.00 kg block shown in the drawing. The magnitudes of the forces are F1 = 55.5 N and F2 = 42.5 N. What is the horizontal acceleration (magnitude and direction) of the block?

    please help break this down for me.

    http://img171.imageshack.us/img171/3181/0411vf7.gif
    You want to start with a Free-Body Diagram. I'm defining a +x axis to the right and a +y axis upward.

    You have two given forces, F_1 and F_2. There are two implied forces, the weight w acting downward, and the normal force N acting upward.

    So take components of each force in the x and y directions and use Newton's 2nd in each component direction. (We automatically know that there's no acceleration in the y direction, right? So all the acceleration must be in the x direction.)

    -Dan
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    okay, so you already have the x component of force 2 right, its the same as the force?

    also using trig i believe the x component of the force 1 angle is 23.5

    and i know accleration is net force / mass

    so couldn't i just add the x components and divide by the mass?

    i'm confused here.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    okay, so you already have the x component of force 2 right, its the same as the force?

    also using trig i believe the x component of the force 1 angle is 23.5

    and i know accleration is net force / mass

    so couldn't i just add the x components and divide by the mass?

    i'm confused here.
    You aren't confused, you have it right! (Just make sure that, in my coordinate system at least, that you have the x component of F2 as negative since F2 is pointing in the negative x direction.

    -Dan

    Edit: You should probably be carrying more digits in your intermediate answers, though, in case you aren't. For example F_{1x} \approx 23.4553~N not 23.5. Using more decimals gives you a slightly better answer at the end, where you can then round off properly.
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    i think i'm close but not correct just yet.

    okay so the x0 of force 2 is: -42.5

    and x0 of force 1 is: 23.4553

    so if i add these together i get: -19.0447

    so then i divided -19.0447 by the mass

    and got about -3.81...

    nevermind i just realised that i can't have a negative acceleration. its right
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    i think i'm close but not correct just yet.

    okay so the x0 of force 2 is: -42.5

    and x0 of force 1 is: 23.4553

    so if i add these together i get: -19.0447

    so then i divided -19.0447 by the mass

    and got about -3.81...

    nevermind i just realised that i can't have a negative acceleration. its right
    A detail that many people tend to forget is that these are vector equations. What we are doing is taking components along one direction or another. For simplicity we define a positive component to be in one direction and a negative component to be in the other.

    Your first answer is correct: The acceleration comes out to be -3.81~m/s^2.

    What does this mean? It simply means that the acceleration is in the -x direction, or as we defined it, off to the left of the diagram.

    -Dan
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