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Homework Help: Finding F1 - Summing forces on a box

  1. Feb 12, 2014 #1
    Finding F1 -- Summing forces on a box

    1. The problem statement, all variables and given/known data

    There are two forces on the 2.80 kg box in the overhead view of the figure but only one is shown. For F1 = 10.9 N, a = 10.2 m/s2, and θ = 34.9°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)

    I'm not sure if this link will work, but here's the picture: http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c05/fig05_32.gif

    2. Relevant equations


    3. The attempt at a solution

    Ok so I've been stuck on this one for a while. I started out finding the sum of the forces:
    ∑F=(2.8kg)(10.2m/s2)=28.56 N

    Then I tried solving for F2:

    Then I found the x and y components of the acceleration:

    Could you multiply ax and ay by the mass and get the F1x and F1y?
    I'm so confused, could someone please help guide me through this problem?
  2. jcsd
  3. Feb 12, 2014 #2


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    Homework Helper

    The x and y components of acceleration are the net accelerations in those directions. Multiplying those by the mass will give you the net forces in the x and y direction.

    If you sum the forces in the x direction, they should add up to the resultant force in the x-direction. A similar exercise will yield the resultant force in the y-direction.

    You need to use this fact to help you find F2.
  4. Feb 12, 2014 #3
    Ok, so I should multiply ax*mass to get ∑Fx? which would give me
    And for the y component: ∑Fy=23.44

    So then would you find F1? To me it looks like F1 only has an x component... So in unit vector notation: 10.9i+0j?

    And then could you solve for F2? Or am I still going about this the wrong way?
  5. Feb 12, 2014 #4
    Your approach to the problem is just a bit off (at least in your first post -- in your second you're working through it better). Remember that Newton II is a vector equation, so you can't use magnitudes for the force or acceleration directly. You can only use magnitudes if you do the equation by component. Set up the equation like this:
    [itex]\Sigma \stackrel{\rightarrow}{F} = m \cdot (\stackrel{\rightarrow}{F}_{1} + \stackrel{\rightarrow}{F}_{2})[/itex]

    Then split that up into components to solve for [itex]F_{2x}[/itex] and [itex]F_{2y}[/itex].
  6. Feb 12, 2014 #5
    I'm sorry, I still don't understand. I tried plugging in values into the formula:
    Is the acceleration vector the same as the F2 vector, or would the acceleration vector be along the ∑F vector?
  7. Feb 12, 2014 #6
    No, for the first part, you're using the wrong values. In the problem statement, you're only given magnitudes and directions for F1 and for acceleration. Whenever you add vectors (like adding F1 and F2), you can't just add magnitudes together, because it's not arithmetic addition, but instead vector addition, so it has to be done by x and y components of the two adding vectors.

    And acceleration will point in the same direction as the net force.
  8. Feb 12, 2014 #7
    Sorry, I put in my equation wrong in my earlier post. It should actually be:

    [itex]\Sigma \stackrel{\rightarrow}{F} = (\stackrel{\rightarrow}{F}_{1} + \stackrel{\rightarrow}{F}_{2}) = m \cdot \stackrel{\rightarrow}{a}[/itex]
  9. Feb 12, 2014 #8
    It almost makes sense! So would my x and y components of the acceleration vector be the same as my x and y components of the ∑F vector?


    Ohh that's ok, that formula makes a whole lot more sense now!

    So can you find the x and y components of F1 vector by just looking at the picture? It doesn't have a y component, right? Isn't it just moving in the x direction? So: 10.9i+0j?
  10. Feb 12, 2014 #9
    Just about, but not quite. The acceleration vector is parallel to net force, but that doesn't mean their the same. Force isn't the same thing as acceleration, right? Look at the equation for Newton II -- net force is a scaled copy of acceleration, so to go from acceleration to net force, you have to multiply by mass. Also, when you found your acceleration components, you got the right magnitudes, but not direction. Both your x and y components are positive, but in the diagram acceleration points in the third quadrant, so both x and y should be negative. Always look at your diagrams to check signs.

    And yes, you can assume the angle between F1 and the x-axis is zero, so the x component is the magnitude, and it doesn't have a y component.
  11. Feb 12, 2014 #10
    Ohh, I can't believe I messed up on the directions of the acceleration vector components!
    Sorry, I'm a little thick headed. Are you saying that I can multiply the acceleration components by the mass and get the components of the net force?
    So: ∑Fx=16.35
  12. Feb 12, 2014 #11
    Yes, that's just a result of the equation [itex]\Sigma F = m \cdot a[/itex] (but remember to check the signs on net force too.) Then, just resolve F1 into its components (as you did), and then you can solve for the components of F2, and then get stuff like the direction and magnitude of it.
  13. Feb 12, 2014 #12
    Ok yay! I got part A. To find the magnitude I tried to just do ∑F=F1+F2
    But I'm pretty sure that's wrong...
  14. Feb 12, 2014 #13
    It's the same mistake again -- that equation can't be used to solve for a magnitude, because it's vector addition. Try instead finding the magnitude using the components.
  15. Feb 12, 2014 #14
    Oh yeah, sorry! Would you just combine the x components?
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