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Where do i start?

  1. Nov 4, 2004 #1
    I have no clue what to do on this problem. it is

    Rachel pulls her 18kg suitcase at a constant speed by pulling on a handle that makes an angle with the horizontal. The frictional force on the suitcase is 27N and Rachel exerts a 43N force on the handle.

    A) what angle does the handle make with the horizontal?

    B) what is the normal force exerted on the suitcase?



    Where do I start?
     
  2. jcsd
  3. Nov 4, 2004 #2
    What does "constant speed" imply?
     
  4. Nov 4, 2004 #3
    Thanks for the help, but i solved this eventually with my lab partner. I missed the lecture unfortunately, but i am up to speed now.
     
  5. Nov 6, 2004 #4
    can you please post answers?? is the normal force 20 N? how did u find the angle?? did you use Tan-1 ( Opp/adj ) ?? if so , what how did u find the opp and adj angle..

    thanks
     
  6. Nov 8, 2004 #5
    isnt normal force = 18g ????? and the angle is cos inverse of (27/43) ?????? what i did is 43cosx = 27 for the angle ....... and mg = normal force. Is this correct ????
     
  7. Nov 8, 2004 #6
    the force 43 N can be separated into the horizontal component (43cosx) and the vertical component (43 sinx).
    Since it is constant motoin, so 43cosx=27, cosx=27/43, x=51.1 degrees.
    But the suitcase is only doing horizonal motion, not vertical motion (it is not moving up and down...), so the two vertical forces (mg ,and the force on the suitcase by the floor) must be cancelled. the downward force: mg=18x9.8=176.4 N which should be equal to the upward force, but now 43sinx=43 sin (51.1degrees)=33.4 N ,which is much less than 176.4N,
    so the unbalanced force 176.4-33.4=143N is the normal force...

    we can also find out the coefficient of friction of the floor since friction=uN, where u is the coefficient and N is the normal force, now the normal force is 143N, so we can know that u=friction/N=27/143=0.19
     
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