Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dynamics - Finding the Net Force

  1. Jun 1, 2005 #1
    Once again, this is probably a very easy question, but I just cant get the answer I am supposed to get.

    The question is:

    A force of 2.0 N east and a force of 3.0 N 25degrees north of east act on an object. What is the net force of the object?

    Now when I follow the system the book uses, I get an answer of 1.46 N 60degrees E of N.

    The answer is apparently 4.9 N 75degress E of N, or 15degrees N of E.

    I don't understand where I am going wrong. Can somebody explain the generic steps I need to go through in order to solve a problem such as this?
  2. jcsd
  3. Jun 2, 2005 #2


    User Avatar

    Anthony, good to see you found this place.

    I stared at this question for a good two minutes before it dawned on me what to do. But I got it in the end lol.

    With these questions always draw a triangle. It's the easiest way to solve them. This particular triangle is a little tricky to get right. But once you draw it you use the cosine law to find the long side of the triangle then you use the sine law to find the angle it's asking for.

    Draw the forces as two "lines" (vectors), but do not draw them from a common point. Draw the first 2N force, then from the end of that line draw the 3N force at an angle of 25 degrees. The line connecting the start of the two lines and the end of the two lines is the net force acting on the object. The angle this long side of the triangle is at is the angle of inclination asked for in the question.

    If you have any more questions just ask. No question is too "dumb".

    (I'm evman150 from CP btw if you hadn't guessed that already.)
  4. Jun 2, 2005 #3


    User Avatar
    Homework Helper

    This is a simple vector addition problem.

    [tex] \vec{A} = 2 \vec{i} [/tex]

    [tex] \vec{B} = 3 \cos(25^{o}) \vec{i} + 3 \sin(25^{o}) \vec{j}[/tex]

    [tex] \vec{C} = \vec{A} + \vec{B} [/tex]

    [tex] \vec{C} = 4.72 \vec{i} + 1.27 \vec{j} [/tex]

    [tex] |\vec{C}| = 4.89 [/tex]

    [tex] \theta_{c} = 15^{o} [/tex]
  5. Jun 2, 2005 #4


    User Avatar

    Generic Steps:

    1. Estimate an answer. Try and think of what your answer is going to look like. In this case, is the answer going to be smaller or larger than the two values given? You should notice that it is going to always be larger, so when you get an answer that is smaller (and when you don't have the answers ;) ), you know you've made an error somewhere.

    2. Draw a diagram. This almost always helps.

    3. Draw a triangle. These types of problems can almost always be solved using some kind of trigonometry.

    4. Make sense of the triangle. Draw the proper forces and any angles given. See if you can trigonemtrically find any of the other sides/angles.

    5. Figure out what part of the triangle the question is asking for. No sense coming this far if you're going to screw up now. This is where your initial estimate comes in handy. If you pick the wrong side/angle, your answer will usually be off by a lot and by your initial guess you will be able to catch your mistake and pick the correct answer.
  6. Jun 2, 2005 #5


    User Avatar

    That's beyond his scope. He's done about two months of physics. Sure he could learn that, but 1) there's no use complicating things and 2) I think it's better to learn the fundamentals the graphical/visual way then just doing vector addition.
  7. Jun 2, 2005 #6
    I think it will be better to use the parallelogram method , which goes like:

    If P and R are forces acting at an angle Q , then the resultant force is:

    R= /sqrt P^2 + R^2 +2PRcosQ

    The above equation will give you the reultant and the direction can be easily calculated . Suppose the resultant R makes and angle T with P , then:

    tanT = R cosQ/P+RsinQ
  8. Jun 2, 2005 #7


    User Avatar
    Homework Helper

    Ok sure, Ek, i understand, the problem is when i read Dynamics, I thought it was a dynamic course, instead of a general physics one, although the problem should have hinted me the other way. Yes indeed start with the fundamentals.
  9. Jun 2, 2005 #8
    I am sorry but I still can't figure out how to get 4.9N and 75degrees.

    It's clearly been too long since my last math course... I was under the impression that cosines, sines, and tangents can only be used for right angle triangles?
  10. Jun 2, 2005 #9


    User Avatar

    Cosine Law: [tex]c^2 = a^2 + b^2 - 2ab cosC[/tex]

    Sine Law: [tex]\frac{sinA}{a} = \frac{sinB}{b}[/tex]

    Those are the two laws used to solve non-right triangles.

    I'll draw up a diagram...five minutes.
  11. Jun 2, 2005 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Why don't you describe the method that your book teaches for adding vectors and show us how you arrived at your answer. Then we can see where you went wrong.

    For example: Does you book tell you to break up the vectors into components? Or draw a parallelogram? Etc...
  12. Jun 2, 2005 #11


    User Avatar

    Here's a pretty little diagram.

    Though I do like the parallelogram method posted. I've never come across that before.

    I agree with Doc, it's a good idea to describe to us what you tried or what you've been taught to do and we can build on that. Although I can't help helping out the guy who puts together all the Flames montages, videos etc. :)

    Attached Files:

    Last edited: Jun 2, 2005
  13. Jun 2, 2005 #12
    Ok, here is the system the book used for another example problem:

    http://www.recklesscaution.com/workpage.jpg [Broken]

    I feel foolish for not being able to manipulate that one to work in this situation.

    Thankyou for all the effort so far!
    Last edited by a moderator: May 2, 2017
  14. Jun 3, 2005 #13

    Doc Al

    User Avatar

    Staff: Mentor

    Now we are getting somewhere. The method used in the book is to decompose each vector into x and y components, add those components, and then use the Pythagorean theorem to find the magnitude of the resultant, and a little trig to find the angle that the resultant vector makes with the x-axis. (Note that "East" is considered the +x direction and "North" is the +y direction.)

    [Although you may not recognize it (since it is written in the language of unit vectors) this is exactly what Cyclovenom was explaining.]

    Nonetheless, here are the steps:
    (a) Take each vector and find its x and y component. Draw a diagram for each, then use [itex]F_x = F \cos \theta[/itex] and [itex]F_y = F \sin \theta[/itex]. (Remember: signs matter! If the vector is above the x-axis, then its y-component is positive; if below, negative. Etc.)
    (b) Find the x and y components of the resultant vector by adding the x and y components of the individual vectors
    (c) find the magnitude of the resultant using the Pythagorean theorem: [itex]R^2 = R_x^2 + R_y^2[/itex]
    (d) find the angle the resultant vector makes by drawing a diagram and using the fact that [itex]\tan \theta = R_y/R_x[/itex]. ​

    Now, show us what you did to get your answer, step by step. Start with step (a): What do you get for the components of the two vectors you are trying to add? (Remember: signs matter!)
  15. Jun 3, 2005 #14
    2N east:
    x= 2cos180 = -2
    y= 2sin180 = 0

    3N north of east
    x= 3cos25 = 2.7189
    y= 3sin25 = 1.2679

    Rx = -2 + 2.7189 = 0.7189
    Ry = 0 + 1.2679 = 1.2679

    [itex]R^2 = R_x^2 + R_y^2[/itex]
    = [itex]0.7189^2 + 1.2679^2[/itex]
    = .5169 + 1.6075
    = 2.1243
    sqrt= 1.4575

    [itex]\tan \theta = R_y/R_x[/itex]
    = 1.2679/.7189
    = 1.7635
    = 60.445 degrees

    I must be making one very obvious mistake that I can't figure out.
  16. Jun 3, 2005 #15


    User Avatar
    Homework Helper

    If the vector is pointing east, why do you use an angle of 180 with respect to +x?
  17. Jun 3, 2005 #16

    Doc Al

    User Avatar

    Staff: Mentor

    As Cyclovenom points out, this is wrong. (Where did you get 180 from?) Since the vector points directly along the +x axis (which is East), you shouldn't need to do any calculation to find its x and y components.
  18. Jun 3, 2005 #17
    Ohhh, I assumed that 180 was required as it was a straight line.

    There we go, I get the right answers. I knew it must be something very obvious I overlooked.

    Thankyou once again for all your help!
  19. Jun 3, 2005 #18

    Doc Al

    User Avatar

    Staff: Mentor

    Note that as soon as you posted your work, it only took about a second to spot your error and get you back on track. (That's why it pays to show your work!) :smile:
  20. Jun 5, 2005 #19
    I have another question.

    I am again following an example in the book but end up with an incorrect answer.

    A 125 N box is pulled east along a horizontal surface with a force of 60 N acting at an angle of 42degrees with the horizontal. If the force of friction on the box is 15 N, what is the acceleration?
    The answer is apparently 2.32 m/s[itex]^2[/itex].

    Now, following the past example I go through this solution process:

    cos42 = T[itex]_x[/itex]/60
    T[itex]_x[/itex] = 44.588

    F = T[itex]_x[/itex] - F[itex]_f[/itex]
    F = 44.588 - 15 = 29.588

    a = F/m
    a = 29.588/125 = .236704 m/s[itex]^2[/itex]

    Now comparing my solution process to the example's I can't figure out where I went wrong. If anybody can spot my mistake, it would be very helpful!
  21. Jun 5, 2005 #20


    User Avatar
    Homework Helper

    125 N is the measure of the weight, It's not the mass!.

    Use this to find the mass, where g is 9.81 m/s^2.

    [tex] F_{g} = mg [/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook