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

And here i come again friction

  1. Jun 4, 2005 #1
    so now i'm undestanding a little bit more, but i'm still confused...

    1) a momentum of 96 Nm is applied to the cylinder. The radius of the cylinder is 1 m. [tex] \mu_{A} = \mu_{C} = 0.25 [/tex] The weight of the cylinder is 100 N. What is the minimum vale of [tex] \mu_B [/tex] so the ramp moves to the right?. Weight of the ramp = 400 N. point B is the contact point between cylinder and the ramp.

    My starting system of equations is:

    [tex] 96 \ N = F_{B} + F_{A} [/tex]

    Forces acting on the cylinder:

    [tex] \sum{F_x} = -N_{B}\sin30 - F_{B}\cos30 + N_{A} = 0 [/tex]

    [tex] \sum{F_y} = F_{A} + N_{B}\cos30 - F_{B}\sin30 - W_{cylinder} = 0 [/tex]

    Forces acting on the ramp

    [tex] \sum{F_x} = F_{B}\cos30 + N_{B}\sin30 - F_{C} = 0 [/tex]

    [tex] \sum{F_y} = F_{B}\sin30 - N_{B}\cos30 +N_{C} - W_{ramp} = 0 [/tex]

    From there, i go all the way to:

    [tex] \mu_{B} = 0.55518 [/tex]

    so far so good?, or i'm missing something? please refer to attachment 1

    2) the body A 'weighs' 2500 kg. and body B 'weighs' 500 kg. [tex] \mu_{S} [/tex] between A and B and between B and the ground is 0,2; and between A and the high surface is 0,5. Find the minimum required force P to start the movement of body B to right. Consider all the posibilities.

    Ok, so when they say 'consider all the posibilities' i assumed that they meant that either x = z or x [tex] \neq [/tex] z.

    from there

    forces acting on body A

    [tex] \sum{F_X} = F_{A} = F_{C} [/tex]
    [tex] \sum{F_Y} = W_{A} = N_{A} + N_{C} [/tex]

    forces acting on body B

    [tex] \sum{F_X} = P = F{_B} - F_{A} [/tex]
    [tex] \sum{F_Y} = N_{B} = N_{A} + W_{B} [/tex]

    [tex] \sum{M_{center \ of \ body \ A}}=\left\{\begin{array}{cc}N_{A} = N_{C},&\mbox{ if }
    x = z\\\frac{N_{A}}{N_{C}} = \frac{d_{2}}{d_{1}}, & \mbox{ if } x\neq z\end{array}\right. [/tex]

    [tex] \sum{M_{point \ A}}=\left\{\begin{array}{cc}\frac{W}{2} = N_{C} \ \mbox{(which, from previous balance of momentum)} = N_{A},&\mbox{ if }
    x = z\\N_{C} = \frac{d_{1}W_{B}}{d_{1}+d_{2}}, N_{A} = \frac{d_{2}W_{B}}{d_{1} + d_{2}}& \mbox{ if } x\neq z\end{array}\right. [/tex]

    so in both cases, i end up with P = 981 N, but as you can see [tex] \sum{F_X} [/tex] of body A can not be satisfied if we were to satisfy also the [tex] \sum{M} [/tex], when x = z

    3) i'm really growing fond of this Latex typesetting. I've seen quite a lot software for writing documents with Latex. which one do you recommend the most, for writing papers and/or a thesis?

    thanks in advance for any help you can give me

    Attached Files:

    • 1.jpg
      File size:
      26.2 KB
    • 2.jpg
      File size:
      71.6 KB
    Last edited: Jun 4, 2005
  2. jcsd
  3. Jun 4, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    I haven't looked at #2 yet. #1 still has me going. I thought I agreed with your equations, from which I deduced that point A is at the wall and point C is the floor. I found your result, based on a solution to the first 3 equations using the known coefficient at the wall. The result was

    [tex]F_B = 58.24N[/tex]

    [tex]N_B = 105.49N[/tex]

    from which you might be inclined to conclude

    [tex] \mu_B = \frac{F_B}{N_B} = 0.552 [/tex]

    Unfortunately, those forces calculated from the first three equations are insufficient to move the ramp. If you plug them into the last 2 equations you will find that they are inconsistent.

    This suggests that your first three equations are overspecifying the situation, and you cannot solve for the forces using just those three equations. Two ideas come to mind that need to be explored. The first is that the applied moment is simply not strong enough to move the ramp. Maybe the ramp is just too heavy (too much frictional force at the lower surface) to move it with a moment of 96Nm. The second is that the thing is moving as a result of the applied moment AND that it is accelerating. If the second condition holds then your first equation is not valid. You need to change it to represent an angular acceleration of the cylinder resulting from unbalanced moments, and relate that acceleration to the linear acceleration of the cylinder (downward) and the ramp (to the right). If there is acceleration, your first third and fourth equations all have to change.

    I think the first thing to try is to allow the applied moment (M) to be a variable, assume no slipping at the ramp, and find the minimum M needed to move the ramp. If M must be greater than 96Nm, then there can be no movement of the ramp. If the minimum required M is less than 96Nm, then you know the ramp and cylinder are going to accelerate.
  4. Jun 4, 2005 #3
    hello, olderdan

    thanks for helping me out:

    mhhh, i don't know if what i did is correct, but i added the [tex] \sum{F_X} [/tex] in the cylinder and the [tex] \sum{F_X} [/tex] in the ramp.

    so we have:

    [tex] N_{A} = F_{C} [/tex]

    and doing the same thing with [tex] \sum{F_Y} [/tex] we have

    [tex] F_{A} - W_{cylinder} - W_{ramp} + N_{C} = 0 [/tex]

    so from there i get:

    [tex] N_{C} = 470.58 \ F_{C} = 117.64 \ N_{A} = 117.64 \ F_{A} = 29.41 \ F_{B} = 66.59 \ N_{B} = 119.942 \ \mu_{B} = 0.55518 [/tex]

    thanks for the help, let me continue working with your suggestions
  5. Jun 5, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    That looks better than what I did. I'll see if I can reslove it with what I found, or see where I went wrong.

    I guess the problem is I can't do Algebra. I must have done the thing 5 times and kept making mistakes. My suspicions were correct however, to an extent. I finally wrote the matrix equation and let my calculator do it and reproduced your answer by solving only the first three equations. Those are sufficient for a solution for the "B" forces and their ratio. The reason my original answers were inconsistent with the last two equations is because they were inconsistent within themselves, and I did not check for that. When I got a ratio so close to yours I assumed you had done it the same way I did, which of course you had not. Now that I'm sure of the answer, I can to it by hand perfectly :rolleyes: . Sorry for the misdirection.
    Last edited: Jun 5, 2005
  6. Jun 5, 2005 #5
    no problemo

    hey, don't worry, even the best make some mistakes

    what are your thoughts on number 2?

    thanks for the help
  7. Jun 5, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    I think "all the possibilities" should put your focus on the relationship between [itex]d_1[/itex] and [itex]d_2[/itex]. Your equation

    [tex]\frac{N_A}{N_C} = \frac{d_2}{d_1}[/tex]


    [tex]N_C = \frac{d_1}{d_2}N_A[/tex]

    is valid as long as point A is to the left of center of Body A, which will always be true if it starts the left and the motion is gradual. Also

    [tex] \sum{F_Y} = W_A = N_A + N_C [/tex]

    [tex] \sum{F_Y} = N_B = N_A + W_B [/tex]

    are valid all the time.

    Two possibilities are that sliding occurs at point A before C, in which case

    [tex]P = P_1 = 0.2 N_A + 0.2 N_B[/tex]

    OR sliding occurs at point C, in which case

    [tex]P = P_2 = 0.5 N_C + 0.2 N_B[/tex]

    By the first ratio this is

    [tex]P = P_2 = \frac{0.5d_1}{d_2}N_A + 0.2 N_B[/tex]

    If the first term in [itex]P_2[/itex] is smaller than the first term in [itex]P_1[/itex], then the slipping will occur at C.

    Suppose the initial configuration favors slipping at A, which would be the case if [itex]d_1 > 2d_2/5[/itex]. The slipping results in a decrease in [itex]d_1[/itex] with no change in [itex]d_2[/itex] until the first term in [itex]P_2[/itex] becomes equal to the first term in [itex]P_1[/itex]. On the other hand, if the initial configuaration favors slipping at C because [itex]d_1 < 2d_2/5[/itex], [itex]d_2[/itex] will decrease with no change in [itex]d_1[/itex] until the first term in [itex]P_2[/itex] becomes equal to the first term in [itex]P_1[/itex]. So what is going to happen when the distance ratio makes [itex]P_1 = P_2[/itex]?

    I happen to have Microsoft Word and MathType that allows LaTex conversion, so that is what I use when I do somethign more complicated. A lot of what I do is just reusing and editing what is here. I would have loved to have these tools when I was writing my thesis, which was to say the least a laborious process. I have not used any of the other software besides what I have.
  8. Jun 5, 2005 #7
    i'm really desperate argh!!!

    uhmm from what i understood, if P1 = P2 then sliding would ocurr in both surfaces when d1 = 0.4*d2.

    you know i thought i had it, but nope...

    MAN, it's frustrating, i really thought i had this one

    please refer to attachments for diagram and fbd....

    Weights: block A = 196.2 N (20 kg)
    Beam BC = 245.25 N (25 kg).

    The ramp in which block A lies is flat (i assume that means [tex] \mu_{between \ A \ and \ ramp} = 0 [/tex])

    [tex] \mu_{between \ Beam \ BC \ and \ block \ A} = 0.6 [/tex])

    the 'union' in C is simple and without friction. Calculate the force exerted over the beam in B.

    ok, so i say 'oh, pretty easy, let's do the balance'

    [tex] \sum{F_x} = -176.58(\frac{4}{5}) + N_{A}\frac{3}{5} + F_{BC}\frac{4}{5} - N_{BC}\frac{3}{5} = 0 [/tex]

    [tex] \sum{F_y} = 176.58(\frac{3}{5}) + N_{A}\frac{4}{5} - F_{BC}\frac{3}{5} - N_{BC}\frac{4}{5} - W_{block \ A}= 0 [/tex]

    and i get

    [tex] N_{B} = 98.1 N [/tex]

    [tex] N_{A} = 255.06 N [/tex]

    and this satisfies the system above, but i was curios about 'then why was the beam weight given?' and i did a momentum sumatory in the beam, and the equation i got was:

    [tex] \sum{M_C} = \frac{W}{2} = 0.6*N_{B}*\frac{3}{5} + N_{B}*\frac{4}{5} [/tex]

    and i get that [tex] N_{B} = 105.71 [/tex]


    i supposed that in C the union would exert a reaction, but still i got the same result because

    [tex] \sum{M_B} = F_{Y \ in \ C} = \frac{W}{2} [/tex]

    and then [tex] F_{Y} - W + F_{B}\frac{3}{5} + N_{B}\frac{4}{5} = 0 [/tex]

    so i got the same result again

    that little difference... why does it ocurrs?

    thanks again for your help, really, i appreciate it a lot

    Attached Files:

    • why.GIF
      File size:
      1.8 KB
    • FBD.GIF
      File size:
      3.1 KB
    Last edited: Jun 6, 2005
  9. Jun 6, 2005 #8


    User Avatar
    Science Advisor
    Homework Helper

    Let's think about this a bit more. It's fairly easy to see that if point A is sliding and point C is not sliding that Body A has no motion. To keep body A from moving to the right the force at C has to be equal to the friction force acting at point A, and has to be less than the maximum possible force at C. When d1 becomes equal to .04*d2, the maximum force at C is reached and Body A begins to slide at C. If the coefficients were constant and independent of motion, then these forces should be able to achieve a condition where they are both equal to their maximum possible values, both points would slide, and Body A would move steadily to the right at the speed necessary to maintain the d1/d2 ratio. This is consistent with your understanding of what would happen.

    I don't think you need to incorporate this into your solution to the problem, but in reality the coefficient of kinetic friction is less than the coefficient of static friction, so what I think what would really happen is that if you start pushing Body B slowly to the right you would reach a point where sliding at C begins, which would suddenly reduce the frictional force at point C, making the frictional force at point A less than maximum and point A would stop sliding. When Body A has moved far enough to the right to increase the normal force at C (reducing the normal at A) point C would stop sliding and A would start sliding again. I think the motion would actually be a series of steps with alternate sliding at points A and C.

    I think I see the problem you are having with the new one. It appears that you solved the first two equations assuming the block is moving so that

    [tex] F_{BC} = 0.6*N_{BC} [/tex]

    and used that relationship to reduce the first two equations to 2 unknowns, then solved for them. Now you find the result is inconsistent with the equation for the beam. This is similar to what I did on that other problem, except that in that case there was motion and I had just made an algebra mistake. In this case, the applied force really is insufficient to make the block move. Assume you do not know the ratio of the frictional to normal force at B, and solve all three equations simultaneously. You will find larger values for the normal forces, and the ratio of the forces at B will be less than 0.6. Your last equation needs to be rewritten as

    [tex] \sum{M_C} = \frac{W}{2} = F_{BC}\frac{3}{5} + N_{BC}\frac{4}{5} [/tex]

    I'm not sure what you were asked to find, but your result will show that the net force on the beam at B is not vertical and leans to the right. That means the reaction at C has to lean to the left. Your last equation is justified if you can assume the horizontal component at C is applied to the lower right corner of the beam. If it were applied at the center or the top, you would have to include the moment due to the horizontal components of the forces at B. Since the thickness of the beam is not specified, and the details of the support at C are not apparent, then I think your equation is OK. If there are details you have not included, then you need to consider the beam more carefully.
  10. Jun 6, 2005 #9
    oops, data missing

    yes, i was asked to find the force exerted by the block on the beam at point B.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook