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Homework Help: Velocity of a block on an inclined plane with pulleys

  1. Dec 5, 2009 #1
    Hey all. I'm new to this forum and by the looks of it, a lot of you really know your physics. I was hoping for some help on this problem I can't figure out. I think I haven't fully grasped all the concepts involved in this question.

    1. The problem statement, all variables and given/known data

    Determine the velocity of the 51-lb block A if the two blocks are released from rest and the 30-lb block B moves 2 ft up the incline. The coefficient of kinetic friction between both blocks and the inclined planes is u = 0.10.


    2. Relevant equations

    Force of friction=(u)(Normal force)

    Having trouble putting it all together as you'll see below.

    3. The attempt at a solution

    Basically I think I am close but I am either missing something crucial or just misunderstand a concept. The first idea had in solving this was to just isolate the right side of the incline plane and solve for the tension in the rope. It was laid out like so:

    Tension = Ffriction + Fparallel

    I took all the forces on the block and set it equal to zero. Than solved for Tension. I assumed this was okay to do since there is no acceleration. Also I think the system is at rest after it moves up 2 ft.

    I then looked at the other side of the incline. On this side, there are tension forces directed upwards, a friction force also upward, and Fparallel downward. I used the Work/energy equation to solve for velocity combining all of these. It looked like this:

    (Fnormal)(2ft) - (Ffriction)(2ft) - (Tension)(2ft) = (1/2)(m)(v)^2

    Basically I use this to solve for velocity. The answer I get is not correct. The reason why I didn't show you the numbers I used is that I am sure that I messed up somewhere in terms of concepts.

    Looking at this, it doesn't really make much sense to me. It looks kind of like a jumbled mess. The thing is that I can do these sorts of problems when they're a bit more basic, but I think the fact that it's an incline on 2 sides with pulleys is really messing me up. If someone could explain how to do this, and also explain what's wrong in my thinking I will be so happy and thankful. I promise I won't disappear on replies.

    Last edited: Dec 5, 2009
  2. jcsd
  3. Dec 5, 2009 #2


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    Hi jamesweston, welcome to PF.
    i)when the blocks are released, both are accelerating. The blocks won't stop after moving 2 ft.
    ii)Since locks are connected by a single rope,the tension in each segment is zero.
    iii) When the block A moves by 2 ft up, the block B moves down by 1 ft. So the acceleration of A is twice that of B.
    Identify the forces acting on A and B and find the acceleration of A and B.
    Then using kinematic equation find the velocity of B.
  4. Dec 5, 2009 #3
    Thanks for the quick reply rl.bhat! I've got a few questions about the reply you gave me.

    Does this mean that when I'm doing the free body diagrams for each individual block, I do not factor in the tension at all?

    Can you explain why there's a difference in the distance the blocks traveled? Also, just to point out, block B moves up 2 feet, so block A moves down 1 ft?

    The forces on block A on the relative x axis are the parralel force, friction, and tension? I use these to equal ma? Than use kinematics?

    Thanks for all the help. Hope I'm not pestering you too much.
  5. Dec 5, 2009 #4


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    i) the blocks are not falling freely because of tension caused by the other block.
    ii) The total length of the rope must remain constant. So when the block A moves up through 2 feet. it releases 2 feet rope. Due to this two segments of rope connected to the block extend equally 1 foot each.
  6. Dec 5, 2009 #5
    Thanks so much for your help. One thing I want to point out is I think that you might be mixing up block A and B. If not then my apologies.

    Anyway, here's what I've done. Unfortunately I don't have the right answer yet.

    Block B (Right side block):

    Tension - Fparallel - Ffriction = ma

    The unknowns in this equation are the tension and the acceleration.

    Block A (Left side block):

    Fparallel - Tension - Ffriction = m(0.5a)

    I used 0.5a since this acceleration is just half the acceleration of block b. The two unknowns in this equation are Tension and acceleration.

    Now that I have two equations and two unknowns, I can cancel out the tensions since they are the same, and then proceed to solve for acceleration. After that, I plug in the acceleration into Vf^2 = Vi^2 + 2ad to solve for Vf.

    This is not getting the right answer. My answer is too high. :yuck:

    I feel so stuck rl.bhat. Do you think you could break what I did down? I'm sorry to bother you. You've been a great help so far.

    Also, I was wondering if it would be easier to use work/energy to solve this. This question is part of my work/energy section.

  7. Dec 5, 2009 #6


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    On block A, the forces are
    F parallel - 2*T - F friction = m(0.5a)
    Try this equation.
    It is possible to solve the problem by conservation of energy principle.
  8. Dec 5, 2009 #7
    Still not getting the right answer :frown:. Maybe it is best to approach this from the energy side. Here's what I am doing.

    Block B (right side):

    T - Fparallel - Ffriction = ma

    T - 30sin30 - 30cos30(0.1) = (30/32.2)a [which becomes...]

    T - 17.60 = 0.93a

    Block A (left side):

    Fparallel - 2T - Ffriction = m(0.5a)

    51sin60 - 2T - 51cos60(0.1) = (51/32.2)(0.5)(a) [substitute in for T from other equation]

    41.62 - 2(0.93a + 17.60) = 0.79a

    6.42 - 1.86a = 0.79a

    6.42 = 2.65a

    2.42 = a

    I now put this into the kinematic equation Vf^2 = Vi^2 + 2ad.

    Eliminate the Vi^2 and solve for Vf

    Vf = sqroot[(2)(2.42)(1)] [the 1 is for the distance 1 meter travelled]

    Vf = 2.2 ft/s <-----which is too high. The right answer is 1.56 ft/s.

    Aggh. I'm so frustrated by this problem. Been working on it forever. Do you think you could show me what I'm doing wrong? Thanks again.
  9. Dec 5, 2009 #8


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    T - 30sin30 - 30cos30(0.1) = (30/32.2)a [which becomes...]
    Check this step. It should be
    T - 30*32.2*sin30 - 30*32.2*cos30(0.1) = (30)a [which becomes...]
    Similarly rewrite the equation for A.
    Now try.
  10. Dec 5, 2009 #9
    Oh damn. I was thinking the lb was a unit of force. I will switch it in both sides..

    ... ...

    Hmm strange thing just happened. I switched it so that I multiplied all lbs by 32.2 and left the masses as how they were. I got the same answer 2.42 for my acceleration which gives me the same velocity. Something is still not right.
  11. Dec 6, 2009 #10


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    The acceleration of block A is 0.5*2.42 m/s^2.
    Now find the velocity.
  12. Dec 6, 2009 #11
    This does give me the right answer, but the thing is is that I already factored in the fact that the acceleration is 0.5 earlier on. So why would I do it a second time? Thanks! I think we're really close.
    Last edited: Dec 6, 2009
  13. Dec 6, 2009 #12


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    If you see the solution, you can see that you have found the acceleration of B i.e. a.
    And the acceleration of A is 0.5a.
  14. Dec 6, 2009 #13
    Hmmm...alright then. This is still really confusing me. I didn't show what I had done but basically I did factor in the 0.5 already here:

    Right side: T = 30a + 566.66

    Substitute that in to the left side equation:

    (51)(32.2)sin60 - 2(30a + 566.66) - (51)(32.2)cos60(0.1) = 51a(0.5)

    1422.19 - 1133.32 - 82.11 - 60a = 25.5a

    206.86 - 85.5a

    2.42 = a

    See I already included the 0.5 earlier on. Is there something I'm not getting?

    Do you think you could explain how to do this with energy as I think that it would be better for me. Thanks a lot rl.
  15. Dec 6, 2009 #14


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    (51)(32.2)sin60 - 2(30a + 566.66) - (51)(32.2)cos60(0.1) = 51a(0.5)
    rewrite this equation as
    (51)(32.2)sin60 - 2(30a + 566.66) - (51)(32.2)cos60(0.1) = 51*a'
    where a' is the acceleration of block A which is equal to 0.5*a . Here a is the acceleration of block B. After solving the equation you got the acceleration of block B not A. You are required to find the velocity of the block A.
  16. Dec 6, 2009 #15
    (51)(32.2)sin60 - 2(30a + 566.66) - (51)(32.2)cos60(0.1) = 51*a'

    1422.19 -60a -1133.32 - 82.11 = 51a'

    206.76 - 60a = 51(0.5a)

    206.76 - 60a = 25.5a

    206.76 = 85.5a

    2.42 = a

    v = sqrt{2(2.42)(1)} = 2.2 ft/s

    The correct answer is 1.56 ft/s

  17. Dec 6, 2009 #16


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    For block A accelerstion is 0.5*2.42
    So v^2 = 2*0.5*2.42 = 2.42
    So v = sqrt(2.42) = ?
  18. Dec 6, 2009 #17
    Hmm okay that does give the right answer but the problem I''m having is here:

    206.76 - 60a = 51(0.5a) <----see I've already factored in the 0.5 here. Why would I factor it in again? I don't understand how this works.

    Thanks rl.
  19. Dec 6, 2009 #18


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    OK. You have written 0.5a as the acceleration of B. But what is a?
  20. Dec 7, 2009 #19
    But that is the acceleration of A isn't it? The a is equal to the acceleartion of B, and 0.5a is equal to the acceleration of A. So that whole equation was set to equal 0.5a since a was already the acceleration of B.

    Not sure if that made sense.
  21. Dec 7, 2009 #20


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    By solving the final equation you have found the acceleration of B, which is equal to 2,42 m/s^2. But you required to find the velocity of A. So the accelerstion of A is 0.5*2.42.
    From that you can find the velocity of A.
  22. Dec 7, 2009 #21
    Hmm okay. I'm still having trouble grasping this because I feel like I have already used the 0.5 earlier. But your way is working so what can ya do. Thanks a lot for your help rl.

    Do you think you could lay out a method to do this with work/energy? Thanks again.
  23. Dec 7, 2009 #22


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    1. You know that when one block goes up the incline 2 ft, the other block goes down the incline 1 ft. Using trig you can convert these to changes in potential energy, one positive one negative.
    2. Calculate the normal force for each block and from this the change in thermal energy of the blocks. Note that this is a positive number for both blocks; it is the negative of the work done by friction and represents the heat added to the blocks by friction.
    3. Write the total energy balance equation, the sum of all the energy changes is zero.

    ΔΚ + ΔEthermal + ΔUgravity = 0

    Each of the deltas above has two terms in it, one for each mass. These you have to calculate separately one by one.

    *** Addendum on edit ***
    When you write the kinetic energy changes remember that block B has twice the speed of block A at any time because its acceleration is twice as big.
    Last edited: Dec 7, 2009
  24. Dec 7, 2009 #23
    Thanks a lot kuruman. I just have a couple questions regarding your method if you don't mind.

    I just use mgh to figure these out for their respective blocks? I find the height using trig and multiply it by the mass and gravity, right?

    Okay I'm kind of confused here. Calculating the normal force is no problem. You're saying that I should use the Ffriction = Fnormal(u) and multiply the result by the distance travelled to find the work done by friction? Why is this a positive number? Isn't it working the opposite way?

    I was just wondering so that I can get a better grasp of the concepts here.... why is the sum of all energy changes zero? How come in other questions we set this equation to equal 1/2mv^2 but here it equals zero?

    Thanks a lot kuruman.
  25. Dec 7, 2009 #24


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    The work done by friction, I agree, is a negative number. However, this work appears as heat on the blocks whose temperature is raised. Thus, the thermal energy of the two-block system increases which means that the change is positive.

    Forces, through work, transform energy from one form to another. If you have a "closed" system, no energy escapes to the outside and no extra energy comes in from the outside, so the sum of the changes must be equal to zero.
    I don't know what these "other questions" are and how you used energy conservation, but look at the work-energy theorem. It says

    ΔK = Wnet

    In this case, the net work is
    Wnet = Wgravity + Wfriction
    If I move everything to the left and change signs, I get
    ΔK - Wgravity - Wfriction = 0

    Now the work done by gravity is the negative of the change in gravitational potential energy (Wgravity = - ΔUgravity) and the work done by friction is Wfriction = -f s (where s is the appropriate displacement) so we have

    ΔK - (- ΔUgravity) - (-f s) = 0


    ΔK + ΔUgravity + f s = 0

    That is the sum of all the energy changes that is equal to zero. Note that the friction contribution is a positive number. Of course, since the two-block system starts from rest, the change in kinetic energy is just the final kinetic energy, and you can move the other two terms back to the right side and write

    (1/2)mv2 = -ΔUgravity - f s

    We have gone full circle. What you have done in the past is the same thing we are doing here, except that the kinetic energy term was placed on the left and everybody else on the right side.
    I hope this clarifies things for you.
  26. Dec 7, 2009 #25
    Wow that was wicked. Thanks kuruman! I'm really learning a lot here. I have two little questions left if it is okay with you.

    I'm kind of confused about this. It doesn't really seem intuitive. Can you explain the difference between why friction would do positive work in some cases but negative in others?

    Also, why is the tension is not factored in?

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