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Math Help - Three practicles...

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    Three practicles...

    Three particles, each with a mass of 0.25 kg, are located at ( - 4.0 m, 0), (2.0 m, 0), and (0, 3.0 m) and are acted on by forces, F1=(-3.0N)y, F2=(5.0N)y, and F3=(4.0N)x respectively. Find the acceleration (magnitude and direction) of the center of mass of the system. [Hint: Consider the components of the acceleration.]
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    Quote Originally Posted by babygirl
    Three particles, each with a mass of 0.25 kg, are located at ( - 4.0 m, 0), (2.0 m, 0), and (0, 3.0 m) and are acted on by forces, F1=(-3.0N)y, F2=(5.0N)y, and F3=(4.0N)x respectively. Find the acceleration (magnitude and direction) of the center of mass of the system. [Hint: Consider the components of the acceleration.]
    There is a simple shortcut to a problem like this. I can give you the proof if you like, but quite simply:
    \sum \vec F_{ext} = m_1 \vec a_1 + m_2 \vec a_2 + m_3 \vec a_3 = M \vec a_{CM} (Where M = m_1 + m_2 + m_3, \sum \vec F_{ext} is the net external force on the system, and a_{CM} is the CM acceleration.) This is true even if the masses are not the same.

    So \sum \vec F_{ext} = -3.0 \hat j + 5.0 \hat j + 4.0 \hat i N (Where \hat i and \hat j are the unit vectors in the x and y directions, respectively.)

    Thus a_{CM} = \frac{1}{3 \cdot 0.25 \, kg} \cdot (4.0 \hat i + 2.0 \hat j) \, N = \left ( \frac{16}{3} \right ) \hat i + \left ( \frac{8}{3} \right ) \hat j \, m/s^2

    So the magnitude of a_{CM} is \sqrt{ \left ( \frac{16}{3} \right ) ^2 + \left ( \frac{8}{3} \right ) ^2} \approx 5.9628 \, m/s^2

    and the direction is \theta = tan^{-1} \left ( \frac{\left ( \frac{8}{3} \right )}{\left ( \frac{16}{3} \right )} \right ) \approx 26.565^o and is in Quadrant I.

    Thus the acceleration of the center of mass is (to the correct number of sig. digs.) 6.0 m/s2 at an angle of 27 degrees above the positive x-axis.

    -Dan
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