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Magnitude of acceleration of boxes -- is this method correct?

  1. Mar 1, 2017 #1
    1. The problem statement, all variables and given/known data

    nkkf0lM.png

    Two boxes with masses [itex] m_1 = 4 kg [/itex] and [itex] m_2 = 10 kg [/itex] are attached by a massless cord passing over a frictionless pulley as shown in in my picture. The incline is frictionless and thetha = 30 degrees
    2. Relevant equations
    f = ma

    3. The attempt at a solution

    Tension 1 is above mass 1 and tension two is to the right of mass 2
    So I find [itex] F_gm1 = 9.8(4 kg) = 39.2 N [/itex] that means [itex] t_1 = 39.2 N [/itex]

    Now on mass 2 I have 3 forces acting on it. Normal force perpendicular to the surface, force of gravity going straight down, and tension force connecting the rope and box.

    [itex] F_g = (9.8)(10 kg) = 98 N [itex] and this force has components:[itex] F_g\cos\Theta [/itex] perpendicular to surface of contact, = to my normal force, and [itex] F_g\sin\Theta [/itex] parallel to my surface. This force gives me the tension in the rope.

    Is what I am saying here correct? It makes sense to me since [itex] F_g\sin\Theta[/itex] is parallel in the opposite direction of tension, and [itex] F_g\cos\Theta [/itex] is parralel opposite direction normal force.

    So I calculate [itex] F_g\cos\Theta = 84.87048957 N, F_g\sin\Theta = 49 N [/itex]

    These components make sense because squaring both, then adding, then taking the square root gives me my [itex] F_g = 98 N [/itex]

    So that gives me [itex] F_n = 84.87048957 N [/itex]

    Tension = [itex] F_t = 49 N [/itex]

    So I think I've dealt with all the forces here. The next part of my question asks:

    What is the magnitude of the acceleration of the boxer?

    So I know f = ma, and the box on the right has a tension of 49 N, and the box on the left has a tension of 39.2 N. So wouldn't my Net forces be 49 N - 39.2 N, going in the same direction as [itex] F_g\sin\Theta [/itex] ? Since that has a larger magnitude (49) opposed to (39.2) the force of gravity on box 1?

    So net force = 9.8 N going in the direction of [itex] F_g\sin\Theta[/itex], now I can use f=ma

    So my question here is, is the m in that equation = to both of the masses that are being involved??

    so would it be 9.8 = 14a, 9.8/14 = a

    would that be the magnitude of acceleration?
     
  2. jcsd
  3. Mar 1, 2017 #2
    You are setting the tension of the rope attached to the 4kg block equal to the weight of the 4 kg block. If that was true, how could the 4 kg block accelerate? Another question to ask yourself: Is there anything that would cause the tension in the rope at m1 to be any different than the tension in the rope at m2? The pulley was specified to be frictionless.
     
  4. Mar 1, 2017 #3
    Well wouldn't it be able to accelerate because there is another tension, [itex] F_t2 = F_g\sin\Theta = 49 N [/itex] on the other side of the rope pulling the weaker, 39.2 N tension?
     
  5. Mar 1, 2017 #4
    It is impossible for an object to accelerate if the sum of the forces equals 0. If the force from the rope acting on the 4 kg block is equal to the weight of the 4 kg block, it simply will not accelerate. Likewise, if it is accelerating, the tension in the rope cannot be equal to the weight of the 4 kg block. The tension at one end of the rope cannot be different than at the other end of the rope unless there is something in between those two points which causes a difference - a pulley with some moment of inertia and friction, or even a rope that has weight, for example.
     
  6. Mar 1, 2017 #5
    Are all of my other forces correct?

    [itex] F_{gm1} = 4 kg (9.8) = 39.2 N [/itex]

    [itex] F_{gm2} = 10 kg (9.8) = 98 N [/itex]

    [itex] F_{nm2} = F_{gm2}\cos{30degrees} = 84.87048957 N [/itex]

    That only leaves me with the sin component of my [itex] F_{gm2} [/itex] force, [itex] F_{gm2}\sin{30 degrees} = 49 N [/itex] that is not being balanced out, so would that be my tension then?
     
  7. Mar 1, 2017 #6
    Since the problem specifies that the cord is massless and the pulley is frictionless, that tells you that the tension in the cord has to be the same across the entire cord. So you could call that force T, for tension. So you can then sum the forces on each block and solve for the force T and the acceleration.

    Yes, all of your forces look correct: 39.2 N, 98 N, and 84.87 N. It even looked like you got the correct answer for acceleration. I am just trying to help you understand exactly what is happening in the problem. Specifically for this problem, you cannot say that the tension at one point in the rope is different than the tension at another point in the rope.
     
  8. Mar 1, 2017 #7
    Okay I think I'm beginning to understand. The point you're trying to get across is: there can only be one tension inside the rope, yes? That makes sense to me since it is being pulled on from two sides, so its going to meet somewhere in the middle with regards to the force being pulled from both sides, so there is only one tension.

    My next question is: What is the tension in the cord connecting the boxes?

    My thinking makes sense to me: It is being pulled 49 N to the left, and 39.2 N to the right. 49 N> 39.2 N, so wouldn't tension in the rope be 49-39.2 N? My book gives me answer 42 N which I don't see how they did it.
     
  9. Mar 1, 2017 #8

    haruspex

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    No. The 49N is the downslope component of the gravitational force acting on the block. It does not directly act on the rope. The block pulls on the rope. You can think of some of the 49N as being used to accelerate the block, and only what is left over reaches the rope.
    But the "right" way is to draw a free body diagram of the block, let the tension be T, and write out the ΣF=ma equation for the block. Do the same for the other block and eliminate the Ts.
     
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