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Answer Part C) only The figure represents a vertical cross section of a smooth block...

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roshan-rox | Valedictorian

Posted August 28, 2013 at 9:06 AM via web

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Answer Part C) only

The figure represents a vertical cross section of a smooth block of mass M with PQ and RS inclined at an angle Alpha  to the  horizontal, and QR and PS being horizontal. A smooth light inelastic string passes  over  two small smooth pulleys at Q and R . Two small smooth particles A and C of masses m_1 and m_3 respectively are attached to the ends of the string. A third small smooth particle B of mass m_2 is attached to the string in between Q and R .The block is free to move on a smooth horizontal plane.

A)

Write downequations to determine the acceleration of the block relative to the plane, the accelerations of  the  particles relative to the block, and the tensions in the portions AB and BC of  the string.

B)

If mass of the particle B is negligible, show that the tensions in the two portions AB and BC of the  string are the same .

C)

If, further the mass of the block is also negligible, show that each of the magnitudes of the  reactions of the block on A and C is equal to `(2m_1m_3)/ (m_1+m_3)g  cos alpha` .

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llltkl | College Teacher | Valedictorian

Posted August 28, 2013 at 4:46 PM (Answer #1)

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With respect to the block, the two particles A and C (assuming B to have a negligible mass) will behave as a single inertial mass, having reduced mass=`(2*m_1*m_3)/(m_1+m_3)`

So, the magnitudes of the normal reactions of the block on A and C will be the same.

According to the attached figure,

<DAE=<GCF=`(90^o-(90^o-alpha)=alpha`

Considering y-component of motion, of A (or C),

`N-(2*m_1*m_3)/(m_1+m_3)*g*cosalpha=0`

`rArr N=(2*m_1*m_3)/(m_1+m_3)*g*cosalpha=0`

Hence the proof.

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