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Deriving Expression for Block Configuration

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See attached file.

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 Summary

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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 provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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