1. Not finding help here? Sign up for a free 30min 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!

IM NEW HEREPLEASE HELP (kinda long)

  1. May 18, 2005 #1
    Hey everyone ..im new here but i really need help on this physics problem so i would appreciate it if someone would help me..

    I kno it would help to have a picture but please try to imagine it..its a connected mass problem on an incline plane

    Starting from rest, it takes 4 seconds for block B..which is m2..(the one hanging) to reach the ground 2 meters belown..what is the coefficient of friction needed between block A..m1 (the one on the incline) and a 30 degree incline plane so that this will occur....block B= 6 kg.....block A= 8 kg

    PLEASE HELP ME IF U KNO..THANX

    im thinkin i would have to find the acceleration first so would it be 0.125m/s/s because a=v/t.....and v=d/t

    v=d/t...2meters/4seconds=0.5m/s and then since a=v/t...0.5m/s divided by 4 seconds is 0.125 m/s/s which is acceleration..well thats what i think..am i off or is this rite???? then how do i find the coefficient???
     
  2. jcsd
  3. May 18, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    d/t will give you the average speed, not the final speed. Since the motion is uniformly accelerated, the average speed is [itex](v_i + v_f)/2 = v_f/2[/itex], so the final speed is 1 m/s. Thus a = v/t = 1/4 = 0.25 m/s/s.

    You can of course also use the kinematic equation [itex]d = 1/2 a t^2[/itex] to solve for a.

    Now that you have the acceleration, you can analyze the forces acting on both masses using Newton's 2nd law. (Draw a careful diagram of all forces acting.) The only unknown will be the force of friction. Once you have that, recall that [itex]F_f = \mu N[/itex].
     
  4. May 18, 2005 #3
    Since the motion is uniformly accelerated, the average speed is , so the final speed is 1 m/s. Thus a = v/t = 1/4 = 0.25 m/s/s. <---can u please explain how u got the final speed to be 1 m/s??

    are u sure [itex]F_f = \mu N[/itex]
    works for inclined planes too??
     
  5. May 18, 2005 #4

    OlderDan

    User Avatar
    Science Advisor
    Homework Helper

    When acceleration is constant, the velocity changes by the same amount in every equal time interval. In this problem, the accelertion was constant for the 4 seconds of interest. The average velocity ina ll cases is the total distance travelled divided by the total time. That is the velocity you calculated. But when acceleration is constant, average velocity is also the average of the initial velocity (0 in this problem) and the final velocity. When you average the final velocity with zero initial velocity you get half the final velocity, so the final velocity is twice the average velocity, or 1 m/s.

    The force of friction still depends on the normal force on an inclined plane, but on an incline the normal force is not equal to the weight of the object. You have to find the component of the weight perpendicular to the plane; the normal force will have the same magnitude as this component. You will also need the component of the weight parallel to the plane to do this problem.
     
  6. May 19, 2005 #5
    For some reason im still confused on how u got 1 m/s ..how did u determine what the final velocity was..did u use the equation VF=Vi + at?? if yes..then how did u get acceleration??? because if you used the equation for acceleration which is v/t..u need the velocity which u dont have..your a great help but can u please break it down more..thanx
     
  7. May 19, 2005 #6
    Would the weight of the component perpendicular to the plane be m2..the one hanging?? and then i would have to find the component of m1 which would be the on one the incline plane??
     
  8. May 19, 2005 #7

    Doc Al

    User Avatar

    Staff: Mentor

    For uniformly accelerated motion, the average velocity can be found in two ways:
    (1) [itex]V_{ave} = D/T[/itex], this tells us that the average speed during the 4 second fall is 2m/4s = 0.5 m/s.
    (2) [itex]V_{ave} = (V_i + V_f)/2[/itex], which just says that the average speed for any interval is right in the middle between the initial speed and the final speed. In this interval, we start from rest so [itex]V_i = 0[/itex], thus we know that [itex]V_{ave} = (V_f)/2[/itex]. Since we already figured out the average speed using method (1), we can use it to figure out [itex]V_f = 2 V_{ave} = 2 (0.5) = 1[/itex] m/s.

    Of course, if you know the kinematic formula [itex]d = 1/2 a t^2[/itex] you can solve for a immediately.
     
  9. May 19, 2005 #8

    Doc Al

    User Avatar

    Staff: Mentor

    The magnitude of the normal force will equal the perpendicular component of the weight of the mass on the incline (m1). That weight acts down; use some trig to find the components of that force parallel and perpendicular to the incline.
     
  10. May 19, 2005 #9
    THANK YOU SOOOOO MUCH..u helped a lot...one more question though..lol if i need to find the coefficient of friction ..u said i have to use this [itex]F_f = \mu N[/itex] rite...but if i plug in Fn what would i plug in for Ff to find [itex]\mu[/itex] ??
     
  11. May 19, 2005 #10
    wait a minute..i think i got it..
     
  12. May 19, 2005 #11
    nvm i dont got it...can i use m2g-[itex]\mu[/itex]m1g=(m1+m2)a to solve for [itex]\mu[/itex] ?? or would i have to use [itex]F_f = \mu N[/itex] ??
     
  13. May 19, 2005 #12

    Doc Al

    User Avatar

    Staff: Mentor

    The next step is to find the force of friction ([itex]F_f[/itex]). The way to do that is to analyze the forces acting on each mass. Start by listing the forces on each and drawing yourself a diagram with those forces labeled. (Remember that the weight of m1 will have a component parallel to the incline as well as perpendicular.)

    You'll have to apply Newton's 2nd law to both masses.
     
  14. May 19, 2005 #13
    so all that information u gave me would help me solve this problem..i attached a pic of it..
     

    Attached Files:

  15. May 19, 2005 #14

    Doc Al

    User Avatar

    Staff: Mentor

    That's the original problem, which matches your description well enough. Now start identifying the forces that act on each mass.
     
  16. May 19, 2005 #15
    ok so now i got the parallel and perpendicular components for m1 and the perpendicular component for m2....now do i add them?? or what
     
  17. May 19, 2005 #16

    Doc Al

    User Avatar

    Staff: Mentor

    Identify all the forces acting on each mass. Name them. Even better, draw a diagram and label the forces. Start with the hanging mass. Here's a hint: I see two forces acting on the hanging mass. (How many forces act on the other mass?)
     
  18. May 19, 2005 #17
    well i'll tell u what forces i identified so far...i have Fn and Fp acting on m1 (the mass on the incline) and then i have Fn acting on m2..the hanging mass ..im thinkin the other mass that u see is probably Fw...? if thats the other force then where do i go from there?
     
    Last edited: May 19, 2005
  19. May 19, 2005 #18

    Doc Al

    User Avatar

    Staff: Mentor

    I assume Fn is the normal force on m1? If so, realize that that force acts on m1, not on m2. What is Fp?

    Here's the kind of analysis I had in mind. For m2 there are two forces: its weight ([itex]m_2 g[/itex]) which acts down, and the tension in the string (call it T) which pulls up. Since you know the acceleration, you have all you need to solve for the tension in the string. Use Newton's 2nd law.

    Once you find the tension in the string, then it's time to start analyzing m1. (That string tension also acts on m1.)
     
  20. May 19, 2005 #19
    Fn=Force normal..so it only acts on m1??
    Fp=Force parallel......why do i need the tension in the string to find the [itex]\mu[/itex] i thought i was using the equation [itex]F_f = \mu N[/itex] ?? im getting so lost...

    Is the equation to solve for tension...FT-mg=ma?? (FT=Force tension)(mg=Fweight)
    If yes..which mass do i plug in m1 or m2???
     
    Last edited: May 19, 2005
  21. May 20, 2005 #20

    Doc Al

    User Avatar

    Staff: Mentor

    Right. The normal force is the perpendicular force that the incline exerts on the the mass. m1 is the only mass on the incline.
    This problem requires many steps. Yes, you'll use [itex]F_f = \mu N[/itex], but to use it we need to first find [itex]F_f[/itex]. And to find [itex]F_f[/itex] we will apply Newton's 2nd law to m1. And to do that, we need to know all the forces acting on m1. The tension is one of those forces.

    This problem is not easy!

    Yes, but be careful when plugging in numbers. Note that the acceleration (which is downward) should be negative.
    I'll let you figure that out. What mass are we analyzing in this equation?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?