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Math Help - Basic Mechanics - dynamics of a particle

  1. #1
    Super Member Quacky's Avatar
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    Basic Mechanics - dynamics of a particle

    A particle of mass 0.5kg is suspended from a vertical string. By means of another string, a second particle is suspended from the first of mass 0.4kg. A force of 10N is applied to the upper string and the particle moves upwards. Find the tension in the lower string and the acceleration of the system.

    Note: Assume resistances other than those provided, and g, which is to be approximated by 9.8N, are null.

    Answer: Tension 4.44N, acceleration 1.31 m/s^2
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  2. #2
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    Write the net force F_1 on the first particle and F_2 on the second particle in terms of the unknown tension. Then, note that their accelerations are the same, so note that F_i=ma for both particles, and then solve for the tension.
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  3. #3
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    Quote Originally Posted by Quacky View Post
    A particle of mass 0.5kg is suspended from a vertical string. By means of another string, a second particle is suspended from the first of mass 0.4kg. A force of 10N is applied to the upper string and the particle moves upwards. Find the tension in the lower string and the acceleration of the system.

    Note: Assume resistances other than those provided, and g, which is to be approximated by 9.8N, are null.

    Answer: Tension 4.44N, acceleration 1.31 m/s^2
    forces acting on the first mass ... T_1 upward , T_2 and m_1g downward

    net force ... T_1 - T_2 - m_1g = m_1a

    forces acting on the lower mass ... T_2 upward and m_2g downward

    net force ... T_2 - m_2g = m_2a


    adding the two equations ...

    T_1 - (m_1 + m_2)g = (m_1 + m_2)a<br />

    \frac{T_1 - (m_1+m_2)g}{m_1+m_2} = a<br />

    \frac{10 - (0.9)(9.8)}{0.9} = a<br />

    calculate a , then determine T_2 , the tension in the lower string from one of the first two equations.
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  4. #4
    Super Member Quacky's Avatar
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    Extremely helpful; thankyou for your patience and time.
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