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    Deriving Expression for Block Configuration

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    Three blocks are configured as in the attachment. A string is attached from block #1 to block #2 (looped over a pulley), and the angle between the string going from mass #2 to the pulley, with respect to straight downward, is θ (block #2 is hanging there, not touching block #3). An external force F is pushing on block #3.

    (a) Derive an expression for F such that θ does not change with time and block #1 does not move with respect to block #3 (in other words, the whole configuration remains the same as it accelerates to the right). Ignore friction throughout the problem.

    (b) Is it possible for this configuration to remain if M2>M1? Explain your reasoning.

    (c) Now suppose there is a friction between block #1 and block #2, given by F(f)=μ(s)M(1)g. Rederive an expression for F.

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    Solution Preview

    See attached file.

    For three blocks together:
    F = (m1+M2+M3)*a ..............(1)
    where a is the acceleration to be calculated so that system acts as a single entity.
    See free body diagram in attached jpg file.
    For M1 block, along the horizontal direction:
    T = M1*a ........(2)
    For M2 block, in vertical direction:
    T*cos(Q) = M2*g .............(3)
    and in horizontal direction:
    T*sin(Q) = M2*a ......(4)
    From eqn(2) substitute the ...

    Solution Summary

    The problem shows all the workings to arrive at the solutions.