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    A rope with a tension force of 20 N pulls the lower block in the figure. The coefficient of kinetic friction between the lower block and the surface is 0.3.

    The coefficient of kinetic friction between the lower block and the upper block is also 0.3.

    What is the acceleration of the 2.0 kg block?
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    I got 1.76m/s^2... it sounds right but its quite complicated and i want to make sure i' ve got it correct
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    my answer is 1.25 ms^-2




    Proof:


    Take g = 10ms-2


    Equation 1:
    T - (0.3)(1)(g) = (1)(a)
    T - 3 = a

    Equation 2:
    20 - T - (0.3)(3)(g) - (0.3)(1)(g) = (3)(a)
    8 - T = 3a

    Equation (1) + Equation (2) gives:

    5 = 4a
    a= 1.25ms^-2




    PS:
    Notice that while considering in Equation (2) the case of the 2kg, we take the mass to be 3kg
    Also notice that in Equation (2), we have made use of two frictional resistance (the Lower and the Upper slides)
    In the case of the upper slide, the normal reaction is the force acting at that cohesive surface - the weight of the 1kg. That is the force causing the grinding friction, it is what we use as normal reaction to the friction surface
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    (Original post by seb5321)
    A rope with a tension force of 20 N pulls the lower block in the figure. The coefficient of kinetic friction between the lower block and the surface is 0.3.

    The coefficient of kinetic friction between the lower block and the upper block is also 0.3.

    What is the acceleration of the 2.0 kg block?
    Name:  Untitled2.jpg
Views: 17
Size:  21.1 KB
    I got 1.76m/s^2... it sounds right but its quite complicated and i want to make sure i' ve got it correct
    What did u do to get this answer, if u tell us ur method we can help, i cant number crunch rn.
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    (Original post by Onlineslayer)
    my answer is 1.25 ms^-2




    Proof:


    Take g = 10ms-2


    Equation 1:
    T - (0.3)(1)(g) = (1)(a)
    T - 3 = a

    Equation 2:
    20 - T - (0.3)(3)(g) - (0.3)(1)(g) = (3)(a)
    8 - T = 3a

    Equation (1) + Equation (2) gives:

    5 = 4a
    a= 1.25ms^-2




    PS:
    Notice that while considering in Equation (2) the case of the 2kg, we take the mass to be 3kg
    Also notice that in Equation (2), we have made use of two frictional resistance (the Lower and the Upper slides)
    In the case of the upper slide, the normal reaction is the force acting at that cohesive surface - the weight of the 1kg. That is the force causing the grinding friction, it is what we use as normal reaction to the friction surface
    Equation 2 is supposed to be meant for the 2.0 kg mass ONLY, so you cannot (out of sudden) take the mass to be 3.0 kg.
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    g = 9.8 at A Level, its only 10 at GCSE.
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    (Original post by Eimmanuel)
    Equation 2 is supposed to be meant for the 2.0 kg mass ONLY, so you cannot (out of sudden) take the mass to be 3.0 kg.
    In this case you have to look at the combined mass. That's gives you the combined weight which gives the normal reaction acting from the ground.

    You cannot treat the 2kg in isolation, if you draw your lines of forces around the 2kg, you would understand that there is an additional force acting with its own weight, so the normal reaction must balance this pair. Also, when accelerating, it's the entire mass of 3kv that is being pulled by the bottom rope whichever way you look at it.
    I hope you understand now.
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    (Original post by Intel_Imperator)
    g = 9.8 at A Level, its only 10 at GCSE.
    That was me being lazy, hence I said "TAKE g"
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    (Original post by Onlineslayer)
    In this case you have to look at the combined mass. That's gives you the combined weight which gives the normal reaction acting from the ground.

    You cannot treat the 2kg in isolation, if you draw your lines of forces around the 2kg, you would understand that there is an additional force acting with its own weight, so the normal reaction must balance this pair. Also, when accelerating, it's the entire mass of 3kv that is being pulled by the bottom rope whichever way you look at it.
    I hope you understand now.

    I don’t really think you understand the situation. You are mixing stuff to confuse yourself.
    We can treat the 2.0 kg mass in “isolation” and this is what Newton’s 2nd law is telling us. If we want to write Newton’s 2nd law equation for the 2.0 kg mass, we need to find all the forces acting on the 2.0 kg mass. It really depends on the system you choose to analyze. We can proceed to solve the problem in a systematic way without self-confusing oneself as follows:


    I call the 1.0 kg mass and 2.0 kg mass as mA and mB respectively.

    For 1.0 kg mass, we can write Newton’s 2nd law equation in the horizontal direction as

    T μ mAg = mAa ------------(1)

    where T is the tension.


    As for the 2.0 kg mass, the Newton’s 2nd law equation in the horizontal direction is

    FTμmAgμ(mA + mB)g = mBa ------------(2)

    where F is the applied force (20 N) on 2.0 kg mass.


    We can eliminate the tension T in (2) by using (1) and then find an expression for a:

     a=\dfrac{F-\left( 3{{m}_{A}}+{{m}_{B}} \right)\mu g}{{{m}_{A}}+{{m}_{B}}}

    After substituting all the given data, one can arrive a as 1.76 m/s2 using g as 9.81 m/s2.


    We can also write the Newton’s 2nd law equation for the whole system (the combined mass which is really 3.0 kg) by treating the tension as internal forces between the two masses.
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    (Original post by Eimmanuel)
    I don’t really think you understand the situation. You are mixing stuff to confuse yourself.
    We can treat the 2.0 kg mass in “isolation” and this is what Newton’s 2nd law is telling us. If we want to write Newton’s 2nd law equation for the 2.0 kg mass, we need to find all the forces acting on the 2.0 kg mass. It really depends on the system you choose to analyze. We can proceed to solve the problem in a systematic way without self-confusing oneself as follows:


    I call the 1.0 kg mass and 2.0 kg mass as mA and mB respectively.

    For 1.0 kg mass, we can write Newton’s 2nd law equation in the horizontal direction as

    T μ mAg = mAa ------------(1)

    where T is the tension.


    As for the 2.0 kg mass, the Newton’s 2nd law equation in the horizontal direction is

    FTμmAgμ(mA + mB)g = mBa ------------(2)

    where F is the applied force (20 N) on 2.0 kg mass.


    We can eliminate the tension T in (2) by using (1) and then find an expression for a:

     a=\dfrac{F-\left( 3{{m}_{A}}+{{m}_{B}} \right)\mu g}{{{m}_{A}}+{{m}_{B}}}

    After substituting all the given data, one can arrive a as 1.76 m/s2 using g as 9.81 m/s2.


    We can also write the Newton’s 2nd law equation for the whole system (the combined mass which is really 3.0 kg) by treating the tension as internal forces between the two masses.
    There is really no point arguing, especially if you are not prepared to learn.

    The Newton's second law you quoted does not consider any mass in isolation of whatever it is carrying. The resultant force on mass 2kg is clearly dependant on the mass of a what it is carrying (1kg) as they are currently being dragged by F.
    I can't begin to imagine what philosophy you are bringing into this and for whatever personal reasons. This is simple physics.
    Kids studying the motion of elevators are even more enlightened on this.

    The right hand side in your equation should have read (ma + mb) a

    But it's obvious you are not ready to accept your mistake as evident in your last paragraph:
    "We can treat the tension as an internal force"
    It's funny what analogy you are hoping to use here. Clearly, this is an Atwood machine, you can never achieve a solution by lumping up the system because you have two bodies in opposite motion. What makes this unique is that the masses are DEPENDENT on each other. You don't need to be an engineer to understand the dynamics of superposition. Or do you?

    I'm sorry I will have to ignore you from this point. Enjoy
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    (Original post by Onlineslayer)
    There is really no point arguing, especially if you are not prepared to learn.

    The Newton's second law you quoted does not consider any mass in isolation of whatever it is carrying. The resultant force on mass 2kg is clearly dependant on the mass of a what it is carrying (1kg) as they are currently being dragged by F.
    I can't begin to imagine what philosophy you are bringing into this and for whatever personal reasons. This is simple physics.
    Kids studying the motion of elevators are even more enlightened on this.

    The right hand side in your equation should have read (ma + mb) a

    But it's obvious you are not ready to accept your mistake as evident in your last paragraph:
    "We can treat the tension as an internal force"
    It's funny what analogy you are hoping to use here. Clearly, this is an Atwood machine, you can never achieve a solution by lumping up the system because you have two bodies in opposite motion. What makes this unique is that the masses are DEPENDENT on each other. You don't need to be an engineer to understand the dynamics of superposition. Or do you?

    I'm sorry I will have to ignore you from this point. Enjoy
    Why do you look at the speck of sawdust in your brother’s eye and pay no attention to the plank in your own eye?
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    (Original post by seb5321)
    A rope with a tension force of 20 N pulls the lower block in the figure. The coefficient of kinetic friction between the lower block and the surface is 0.3.

    The coefficient of kinetic friction between the lower block and the upper block is also 0.3.

    What is the acceleration of the 2.0 kg block?
    Name:  Untitled2.jpg
Views: 17
Size:  21.1 KB
    I got 1.76m/s^2... it sounds right but its quite complicated and i want to make sure i' ve got it correct

    First, I would apologise to the OP where the posting seems to derail from its intended purpose. I believe you already have the correct answer as around 1.8 m/s2 from MasteringPhysics or Knight physics Textbook.


    (Original post by Onlineslayer)
    There is really no point arguing, especially if you are not prepared to learn.

    The Newton's second law you quoted does not consider any mass in isolation of whatever it is carrying. The resultant force on mass 2kg is clearly dependant on the mass of a what it is carrying (1kg) as they are currently being dragged by F.
    I can't begin to imagine what philosophy you are bringing into this and for whatever personal reasons. This is simple physics.
    Kids studying the motion of elevators are even more enlightened on this.

    The right hand side in your equation should have read (ma + mb) a

    But it's obvious you are not ready to accept your mistake as evident in your last paragraph:
    "We can treat the tension as an internal force"
    It's funny what analogy you are hoping to use here. Clearly, this is an Atwood machine, you can never achieve a solution by lumping up the system because you have two bodies in opposite motion. What makes this unique is that the masses are DEPENDENT on each other. You don't need to be an engineer to understand the dynamics of superposition. Or do you?

    I'm sorry I will have to ignore you from this point. Enjoy

    It seems to me that when you are shown to be wrong, you will also accuse other of being wrong. You can see what other people had done for this question.

    https://www.physicsforums.com/thread...system.222110/

    https://www.physicsforums.com/thread...-block.314301/

    Again, very likely that you would say they have done it wrong instead you.
 
 
 
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