Newton's Laws Problems

Discussion in 'Physics & Math' started by kingwinner, Mar 28, 2006.

  1. kingwinner Registered Senior Member

    Messages:
    796
    I am badly stuck on a few Newton's Laws problems. I hope someone can help me out! Thanks a million!

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    1) A 5 kg block rests on a 30 degrees incline. The coefficient of static friction between the block and incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding
    1a) up the incline?
    1b) down the incline?


    I really have no clue of solving this. Would 1a and 1b have the same answer?

    2) A double incline plane has two boxes of equal masses on opposite sides connected by a rope over a pulley. The left incline is 53 degrees and the right incline is 30 degrees. The coefficient of static and dynamic friction are equal, 0.30 for both boxes.
    2a) Show that the system, when released, remains at rest.
    2b) If the system is given an initial speed of 0.90m/s to the left, how far will it move before coming to rest?


    For 2a, I know that the motion of attempted of the system is to the left beucase the left incline is steeper and the boxes have equal masses. So there are 2 frictional forces, both pointing to the right. Knowing all these, I got that the acceleration is 0.69441m/s^2
    . Then, how can the system remain at rest? Can someone explain? I don't understand.

    For 2b, v1=-0.90m/s, v2=0, a=0.69441. Using these, I found that the system moves 0.58m to the left before coming to rest. But can I use the acceleration that I got in part a? Is this the correct way of doing it?

    3) The radius of the earth is about 6370km, while that of Mars is 3440km. If an object weights 200N on earth, what would it weight, and what would be the acceleration due to gravity, on Mars? Mars has a mass 0.11 that of earth.

    I tried using law of universal gravitation equation and it works. I am wondering if there is any way of solving this problem without the use of the law of universal gravitation beucase this equation is not taught in my course yet.​
     
    Last edited: Mar 28, 2006
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  3. James R Just this guy, you know? Staff Member

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    Hints:

    Draw a free-body diagram showing all the forces on the block. Is the friction force the same in parts (a) and (b)? In particular, consider the direction of that force.



    Bearing in mind that the system is at rest, the friction forces on each block are both directed up the slope. They are the forces keeping the system at rest.

    Remember that the friction force has a MAXIMUM possible value equal to (mu)N. Work out the force down the slope for each block, and calculate the maximum possible friction force for each block. If that maximum is greater than the downward force due to all the other forces acting, then the friction force will be just enough to exactly balance the other forces and stop the system sliding.

    You don't mention whether the blocks are connected by a string or something. If they are, you will need to take the tension forces into account, too.

    No, there isn't.​
     
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  5. Poincare's Stepchild Inside a Klein bottle. Registered Senior Member

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    Come on lad...is it easier to push a box up a ramp or down a ramp?

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    OK...hint to start.

    Draw a 30 degree rt triangle. Draw a box on it. Draw an arrow straight down from the box, and label it 49 n. (force of gravity on a 5 kg block)

    Now break the force into two components. One parallel to the ramp, one perpendicular to the ramp.

    Going down the ramp, the parallel force will help you, going up the ramp you have to overcome it also.
     
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  7. kingwinner Registered Senior Member

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    796
    1) But for both 1a and 1b, the box is at rest and the acceleration is 0, right? And the frictional force points up the slope in both cases because its attempted motion in both cases is sliding down. So wouldn't the applied force required be the same for 1a and 1b?

    2a) Apologies! I missed some information about the question. Edit: the two boxes are connected by a rope over a pulley. So the frictional forces will point to the right for both boxes and I got acceleration = 0.69441m/s^2
    . The acceleration isn't zero, then how can the system remain at rest? This is the part that I don't get...

    2b) Can I use the acceleration calculated in part a to solve this problem?​
     
  8. Rosnet Philomorpher Registered Senior Member

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    681
    For 2b, you'll have to use the dynamic friction coefficient, not the static one.
     
  9. kingwinner Registered Senior Member

    Messages:
    796
    But there are the same, 0.30, right?

    Question 2 is really weird. I got a=0.69441m/s^2 for part a, which means that the system is accelerating in the direction
    , but how is that possible? The left incline is steeper......​
     
  10. Pete It's not rocket surgery Registered Senior Member

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    10,167
    If you're pushing it up the slope, then the friction force points down.
    That's because the net force (apart from friction) is upslope, right?

    So when pushing it upslope, your effort is the only force up the slope, and there are two forces down the slope. What are those two forces?
     
  11. James R Just this guy, you know? Staff Member

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    35,610
    kingwinner:

    No. In part (a) the attempted motion is up the plane. In (b) it is down the plane. Therefore, the friction force points in opposite directions.



    The acceleration is zero, and your value is incorrect because you used the MAXIMUM friction force, rather than the actual friction force.

    You can't. The acceleration in part (a) is zero.​
     
  12. kingwinner Registered Senior Member

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    796
    1) For 1a attempted motion is up the slope, so friction is down the slope.
    For 1b attempted motion is down the slope, so friction is up the slope.

    Got it!

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    2a) Why is the acceleration zero? Can you please show me how you arrive at this conclusion and how you calculate it? I worked out all the algebra and still got a=0.69441m/s^2
    .

    2b) But in this case, the box is moving to the left, and I have to use the coefficient of dynamic friction. And since the coefficient of static and dynamic friction are equal, 0.30 for both boxes, I can use the acceleration I got in part a) a=0.69441m/s^2
    to do the calculation, right? I got an answer of 0.58m to the left, by the way.​
     
  13. James R Just this guy, you know? Staff Member

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    Ok, I'll do 2a for you.

    Resolving forces down the plane (parallel) for the box on the 30 degree slope, we have:

    mg sin 30 - T + f1 = m a1 ... (1)

    For the box on the 53 degree plane, we have:

    mg sin 53 - T - f2 = m a2 ... (2)

    Note: a is initially the same for both, since they are connected by a taut string.

    The friction forces on boxes 1 and 2 are:

    f1 = a mu N1 = a mu m g cos 30
    f2 = b mu N2 = b mu m g cos 53

    where a and b are unknown numbers between 0 and 1 (since the friction might be less than its maximum possible value when the boxes are not sliding).

    Assume the system does not slide. We want to know if this is consistent with a and b both being between 0 and 1. a1=a2=0 gives:

    m g sin 30 - T + a mu m g cos 30 = 0 ...(3)
    m g sin 53 - T - b mu m g cos 53 = 0 ...(4)

    Subtract (3) from (4) and divide through by m g:

    (sin 53 - sin 30) - 0.3 (a cos 30 + b cos 53) = 0

    0.1805 b + 0.2598 a = 0.2986
    b + 1.4393 a = 1.654
    b = 1.654 - 1.4393 a

    If b = 0, then a = 1.149, which is not possible.
    If b = 1, then a = 0.4544, which would be possible.
    If a = 0, then b = 1.654 (impossible).
    If a = 1, then b = 0.2147 (possible)

    So, we can say that the range of a is (0.4544,1), and the range of b is (0.2147,1).

    The actual values of the friction forces will depend on the mass m and the tension T in the string, but we have shown that friction can stop the system from sliding.
     
  14. James R Just this guy, you know? Staff Member

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    While I'm at it, lets look at 2(b)...

    In that case, we have:

    m g sin 30 - T + mu m g cos 30 = -ma ...(1)
    m g sin 53 - T - mu m g cos 53 = ma ...(2)

    Note that the friction is kinetic friction now. In the terms of the previous part, a=b=1. The string remains taut, and so the accelerations of the two boxes have the same magnitude a. We want to find a. Subtracting (1) from (2) gives:

    mg (sin 53 - sin 30) - mu m g (cos 53 + cos 30) = 2ma

    Divide by 2m to get:

    a = (g/2)[(sin 53 - sin 30) - (0.3)(cos 53 + cos 30)] = 1.8517 m/s<sup>2</sup>
     
  15. przyk squishy Valued Senior Member

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    Think about it: You're suggesting that the block on the steeper incline would spontaneously accelerate up the slope?

    Here's a simpler problem to get you thinking: A 1 kg block is lying at rest on the ground. The coefficient of static friction is 0.5. Calculate the friction force and the acceleration of the block.

    You did the calculation to get the acceleration for b) in part a). Your answer's correct by the way.

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  16. kingwinner Registered Senior Member

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    796
    2a) What solid, firm argument can I place to refute the answer a=0.69441m/s^2
    and state that the acceleration is 0m/s^2?

    Is it true that friction can't cause acceleration in the same direction as the frictional force? (as in this case, friciton is acting to the right, so it is not possible for the system to accelerate to the right??)

    But when a person is walking, its free body diagram has three forces, including a foward reaction force in the direction of acceleration? (and that is frictional force, right?)​
     
  17. kingwinner Registered Senior Member

    Messages:
    796
    "Here's a simpler problem to get you thinking: A 1 kg block is lying at rest on the ground. The coefficient of static friction is 0.5. Calculate the friction force and the acceleration of the block."

    Ff=0.5(1)(9.8)=4.9N
    Fnet=-4.9N
    a=Fnet/m=-4.9/1
    =-4.9m/s^2???
     
  18. przyk squishy Valued Senior Member

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    3,203
    Sorry: you fell in the trap :m:
    When have you ever seen a block on the ground accelerate spontaneously?

    The box is at rest, so its acceleration is zero.
    Static friction forces match the other forces applied on the block to make the net force zero, up to a certain maximum.

    If you don't apply a force on the block, there's no friction, and no net force.

    If you apply a force of 1N on the block, there will be a -1N frictional force, and the net force is zero.

    If you apply a force of 4.9N on the block, there will be a -4.9N frictional force, and the net force is zero.

    If you apply a force of 10N on the block, there will be a -4.9N frictional force, and the net force is 5.1N, as the force exceeds the maximum frictional force that can be generated. The box starts to accelerate.

    In general you can't just plug in the coefficient of static friction to find the acceleration, because its an inequality, not an equation.
     
    Last edited: Mar 30, 2006
  19. James R Just this guy, you know? Staff Member

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    35,610


    I've already answered this several times, in detail.

    If you're going to ignore my responses, I won't bother responding to you in future.​
     
  20. kingwinner Registered Senior Member

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    796
    Hi, James R,

    Sorry, I don't mean to ignore your responses! I have read it several times but that's not the way that I usually solve these problems...it is too complicated for me...so I am searching for a simple argument that can refute a=0.69441m/s^2 and prove a=0
     
  21. przyk squishy Valued Senior Member

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    3,203
    Simple: static friction stops objects from moving, but doesn't make them move. Work out what the static friction coefficient would have to be for the block to JUST stay in place. If its less than 0.3, the block won't move.
     
  22. kingwinner Registered Senior Member

    Messages:
    796
    But how do you know that is static friction? There is a possibility that the system MAY be moving (before considering calculations), if I calculated a=0.69441m/s^2 for part a) using the coefficient of friction as 0.30 , how can I know whether this is static or kinetic friction?
     
  23. przyk squishy Valued Senior Member

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    3,203
    You have static friction when you have these 2 conditions:

    #1: V = 0. The speeds are zero and there is no sliding. The system being "released" implies that its initial speed is zero.

    #2: You can set a static friction force that exacly cancels the other forces and gives the system zero acceleration, such that the friction force is less than or equal to the (normal force)*(static friction coefficient).

    Simple example: Imagine a block of weight 10 N at rest, and a coefficient of static friction of 0.1. 10 N * 0.1 = 1 N right? This does not give the friction force acting on the block. This is the maximum static friction force that can act on the block. If the net force acting on the block is less than 1 N, the static friction force will balance it exactly so you get no acceleration. If the net force is more than 1 N, you get acceleration, and you have to work out the friction force using the kinetic friction coefficient.

    You never never ever calculate an acceleration using the coefficient of static friction. The only function of the static friction coefficient is to determine whether the system at rest will stay at rest or not.
     

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