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Calculating horizontal acceleration?

  1. Mar 2, 2015 #1
    1. The problem statement, all variables and given/known data
    The problem reads: Two forces act on a 5.0 kg block on a friction-less surface.

    (a) Draw a free-body diagram
    (b) Determine the magnitude of the normal force (Fn)
    (c) Determine net horizontal force
    (d) Determine the magnitude and direction of the horizontal acceleration

    I don't understand how to calculate the direction of the horizontal acceleration...usually for direction I am used to using
    θ = arctan (∑FY/∑FX), however for this problem do I just completely ignore the Y net forces since it is asking for horizontal? Therefore is it just θ = arctan (∑Fx)?


    I was wondering if anyone could also please check my work? Criticism and tips for my free-body diagram would be much appreciated because I struggle with those.
    (a) I1gk2Gv.png

    (b) For determining the normal force because I assume acceleration is 0 in the vertical position
    ∑Fy = Fn - mg sin10° = 0
    Fn = mg sin10°
    Fn = (5 kg)(9.80 m/s^2) sin10° = 8.5 N

    (c) For determing the net horizontal force I calculated the forces F1 and F2 and added them up
    F1: 50 N cos10° = 49 N
    F2: 15 N cos 180° = -15 N
    ∑Fx = 49 N + (-15 N) = 34 N

    I'm not quite sure why I had to calculate F1 and F2 forces using trigonometry when they're already given in the problem...however, I followed this step from a similar homework problem that required that. I'm not sure why this step is required, I merely assumed that's what I had to do so I feel iffy about this part.

    (d) For determining the magnitude of the horizontal acceleration I used ∑Fx = m ax and solved for ax = ∑Fx / m
    ax = (34 N) / (5 kg) = 6.8 m/s^2

    For determining the direction of the horizontal acceleration I'm not sure what to do....
    I suppose it's θ = arctan (∑Fx) ? I really don't know, any help would be much appreciated. Also I suppose the direction is NW?


    Thanks for taking the time to help out.



    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 2, 2015 #2
    First of all, the direction of the normal force Fn is incorrect. It should always be "perpendicular to the contact surface".
     
  4. Mar 2, 2015 #3

    haruspex

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    ... Secondly, where did the 10 degrees come from?
     
  5. Mar 2, 2015 #4
    OH SHOOT lol! Sorry! For all my calculations I meant 30 degrees! No idea why I was putting 10 o_O I guess I was mixed up with a previous problem.


    I always thought the normal force was perpendicular to the x-axis so I made it perpendicular to the F1 force...I always have trouble drawing the normal force and the weight. I always notice the normal force and the weight in many free-body diagrams are slightly off from each other and never a straight line down.
    Is the only problem with my free-body diagram is just the normal force? Should it be straight up from the weight?
     
    Last edited: Mar 2, 2015
  6. Mar 2, 2015 #5

    haruspex

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    As Ashu2912 posted, the normal force is normal to the contact plane. That is not necessarily vertical, but in this case it is. Other than that your FBD looks fine.
     
  7. Mar 3, 2015 #6
    I recommend drawing a compass rose to define x, y, +, - directions as they apply to your answers...

    One way to think of the normal force is to think of the ground pushing up on the object. For an object not to fall through the floor, the floor has to push up on the object as much as gravity is pushing down. On a flat surface this is mass × gravity or weight, on an incline it is different ⇒ FN = mgcos(θ) as weight components are distributed differently ... Think of a bathroom scale on an angle... As you know the reading will be different than it would if was on a flat surface.

    Here is a good link that deals with inclines, normal force, and FBD...

    http://www.physicsclassroom.com/class/vectors/Lesson-3/Inclined-Planes
     
    Last edited: Mar 3, 2015
  8. Mar 3, 2015 #7

    BvU

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    Hello Lex, a (belated) welcome to PF :smile: (and welcome back to the educational world, I might add)

    So the 30 degree issue is settled. Coordinate system is clear from your post #1: positive y is up, positive x is to the right.

    In the FBD, FN is straight up, and now we're ready to apply the relevant equations (oops, where are they gone :wink:)...

    On to the attempt at solution, mistakenly posted under 1. (oops 2):

    In (b), you do the right thing: block doesn't fall through the floor and doesn't fly off, so vertical forces balance. Newton 1 says: no acceleration if no net force in the vertical direction (not: position!). You mention mg and FN. But it seems you forget one: the vertical component of F2. So back to the expression, this time with three contribuants.

    In (c) you do the right thing again, just correct the angle. ANd you have the right sign.

    And the sign is all they want in part (d). Positive, so to the right. This wasn't asking for some real value, just left or right.

    Goes to show physics isn't all that hard ! -- at least not this time :smile:
     
  9. Mar 10, 2015 #8
    Thanks for the replies everyone! So for when it asks for direction it does not want a degree, just left or right? Okay.

    I attempted this problem again, and for the first part I totally forgot why I assigned my mg as sin and not cosine. If anyone could get me back on track on what I was initially thinking that make me choose sine instead of cosine. I feel its cosine because W is on the adjacent side of the 30 degree angle.
     
  10. Mar 10, 2015 #9

    haruspex

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    A few things have changed since your initial post. Please lay out your working as it is now.
     
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