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The figure represents a vertical cross section of a smooth block of mass M with PQ and...

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roshan-rox | Valedictorian

Posted August 28, 2013 at 9:05 AM via web

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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.

Write down equations 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.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted August 28, 2013 at 9:26 AM (Answer #1)

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The figure depicts a block B that is free to move in the horizontal direction. To the block are attached two other blocks A and C of mass `m_1` and `m_3` .

The horizontal motion of block B is influenced by the horizontal components of the forces exerted by block A and block C. These are `m_1*g*sin alpha` and `m_3*g*sin alpha` respectively. The two forces are in opposite direction. The magnitude of the net force is `|m_1*g*sin alpha - m_3*g*sin alpha|` and the resulting acceleration of block B is  `|m_1*g*sin alpha - m_3*g*sin alpha|/m_2` .

The tension in the string is the same throughout and equal to `|m_1*g*sin alpha + m_3*g*sin alpha|`

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llltkl | College Teacher | Valedictorian

Posted August 28, 2013 at 5:09 PM (Answer #3)

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Consider acceleration a of the particles relative to the block, as shown in the attached figure. The equations defining motion of the particles relative to the block would be:

1. `m_1gsinalpha-T_1=m_1a` (particle A, x-direction)

2. `T_1-T_3=m_2a` (particle B, x-direction)

3. `T_3-m_3gsinalpha=m_3a`` `

To find the acceleration, a

----------------

1+3

-----------------

`-(T_1-T_3)+(m_1-m_3)gsinalpha=(m_1+m_3)a`

Plugging in the value of `(T_1-T_3) ` from 2

`-m_2a+(m_1-m_3)gsinalpha=(m_1+m_3)a`

`rArr a=((m_1-m_3)gsinalpha)/(m_1+m_2+m_3)`

`T_1=m_1gsinalpha-m_1a`

(where `a=((m_1-m_3)gsinalpha)/(m_1+m_2+m_3))`

and `T_3=m_3a-m_3gsinalpha `

(where `a=((m_1-m_3)gsinalpha)/(m_1+m_2+m_3))`

Due to this unbalanced force, the block will have an acceleration relative to the plane (their junction is frictionless) and the magnitude can be found out from Newton's third law of motion as:

`A=-((2m_1*m_3)/(m_1+m_3)cosalpha*a)/M`

(Where M is the total mass of the system)

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