1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force Problem

Tags:
  1. Dec 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Barney is practicing his sweeping techniques for an upcoming curling tournament. He exerts a force on a 1.1-kg broom as he walks across a tile floor at a constant speed. The coefficient of friction between the floor and the broom is 0.45 and the broom handle makes an angle of 41° with the horizontal. Determine the amount of force with which Barney pushes downward (along the handle of the broom) in order to achieve this constant speed motion. Begin with a free body diagram. (ans: 10.6N)

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    I created a FBD and using the info from the question I came up with
    (10.78N)+(Fsin41°)=Fnorm
    Fn=11.44N
    (0.45)(Fn)=Fcos41°
    F=6.82N now if this is my horizontal force then it will be the same as my friction force since the broom isn't accelerating.
    If I want the force on the broom then using SOH CAH TOA. I take A/cos41 = 9.03N. I'm at a loss of how to find 10.6N
     
  2. jcsd
  3. Dec 23, 2016 #2

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think you are correct until the end. You have:

    ##0.45F_n = F \cos(\theta)##

    What did you do after that?

    Correction: how did you get ##F_n = 6.82N##? That can't be right.
     
  4. Dec 23, 2016 #3
    Well if (10.78)+(sin41)=11.44N then
    (0.45)(11.44)=Fcos41
    5.148=Fcos41 divide both sides by cos41
    F=6.82N
     
  5. Dec 23, 2016 #4

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    How do you know the normal force is ##11.44N##?
     
  6. Dec 23, 2016 #5
    I thought that if I put the (10.78)+(Sin41) in my calculator that I have my Fn now?
     
  7. Dec 23, 2016 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Where did the ##F## disappear to? It should be ##F \sin\theta##
     
  8. Dec 23, 2016 #7
    Oh woops! Ok so now if I sub (10.78N)+(Fsin41°)=Fnorm
    (0.45)(Fn)=Fcos41°
    (0.45)(10.78N)+(Fsin41°)=Fcos41°
    But now if I move the Fsin41 over to Fcos41 won't that cancel out my F?
     
  9. Dec 23, 2016 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The coeff of friction applies to all of ##F_n##. In any case, you just need a bit algebra now.
     
  10. Dec 23, 2016 #9
    (0.45)(10.78N)+(Fsin41°)=Fcos41°
    4.851N=Fcos41°-Fsin41° won't this cancel out my F's though? Meaning I have no variable anymore?
     
  11. Dec 23, 2016 #10

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's still wrong. The coeff of friction applies to the ##F\sin\theta##. In any case, I don't see why ##F## would cancel out of the equation.
     
  12. Dec 23, 2016 #11
    Sorry, I'm a bit lost. I'll try to show you what I think you mean.
    (0.45)(10.78N)+(Fsin41°)=Fcos41°
    4.851+(Fsin41°)=Fcos41°
    4.851+(F(0.656))=F(0.755)
    5.507N=F(0.755)
    7.29N = F
    I'm not sure what I'm doing wrong.
     
  13. Dec 23, 2016 #12
    The first line is right, but the second line is wrong. Just an algebra error.
    It should be: (0.45)(10.78) + (0.45)(Fsin41) = Fcos41
     
  14. Dec 23, 2016 #13

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You were doing all right in your first post. You got:

    ##F_n = mg + F \sin\theta##

    ##F_f = 0.45F_n##

    Where ##F_f## is the frictional force. And:

    ##F_f = F \cos \theta \ ## (for equilibrium)

    You had all that. But, then you went wrong. You should have got:

    ##F \cos \theta = 0.45F_n = 0.45 ( mg + F \sin\theta)##

    You just applied the ##0.45## to the first term in the normal force. You must apply it to all of the normal force.

    To finish this off, you can in fact just plug in all the numbers and you'll get a numeric equation for ##F##.
     
  15. Dec 23, 2016 #14
    Ok I understand my error with the co-efficient of friction now. But now I'm getting 6.82N again...sorry, I'll show you what I'm doing
    (0.45)(10.78) + (0.45)(Fsin41) = Fcos41
    4.851 + 0.295 = Fcos41
    5.15 = Fcos41 if I divide both sides by cos 41 I get 6.82 again
    6.82N = F
     
  16. Dec 23, 2016 #15

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You've just dropped one of the ##F's## out of the equation. I've highlighted it in bold above.
     
  17. Dec 23, 2016 #16
    I'm sorry this just doesn't seem to be clicking.
    (0.45)(10.78) + (0.45)(Fsin41) = Fcos41
    4.851 + 0.295F = Fcos41
    -0.295F = 0.705Fcos41
    4.851 = 0.705F cos41 divide both sides by .705 then cos 41 and I get 9.12N
    6.88 = Fcos41
    9.12 = F
    I hope I'm at least going in the right direction and I really appreciate both your help!
     
  18. Dec 23, 2016 #17

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I'm not sure why you calculated ##sin(41)## but not ##cos(41) = 0.755##. So, you should have:

    ##4.851 + 0.295F = 0.755F##

    Can you finish it off from there?
     
  19. Dec 23, 2016 #18
    Oh wow that was a bit stupid on my part! It makes sense now! Thanks for all your help, really appreciate it!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Force Problem
  1. Force Problem (Replies: 3)

  2. Forces problem (Replies: 6)

  3. Force monkey problem (Replies: 5)

  4. Force problem (Replies: 9)

Loading...