(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

19: There is a block a, 100kg and block b 10kg on the right towards the top of block a, hanging in mid air. a force is applied to these two blocks on the left, pointing to the right. THe problem is as follows: What force must be exerted on block A in order for block B not to fall? First, draw a Free Body Diagram of each block and one of them together. The coefficient of static friction between blocks A and B is 0.55, and the horizontal surface is frictionless.

2. Relevant equations

[tex]

\sum F = ma

[/tex]

[tex]

F_f = \mu F_n

[/tex]

3. The attempt at a solution

I started off by finding the applied force in terms of the acceleration of the system in the x direction.

[tex]

\sum F_a = m_a a

[/tex]

[tex]

m_a a = F_a - F_c

[/tex]

[tex]

\sum F_b = m_b a

[/tex]

[tex]

m_b a = F_c

[/tex]

Therefore

[tex]

F_a = a ( m_a + m_b )

[/tex]

Then I substituted in for acceleration

[tex]

a = \frac {F_c}{m_b}

[/tex]

[tex]

F_a = \frac {F_c ( m_a + m_b )}{m_b}

[/tex]

After that I moved to the y direction to solve for the contact force

[tex]

\sum F_b = 0

[/tex]

[tex]

0 = F_f - m_b g

[/tex]

[tex]

m_b g = \mu F_c

[/tex]

[tex]

F_c = \frac {m_b g}{\mu}

[/tex]

I then substituted that back into my equation for the x direction

[tex]

F_a = \frac {m_b g (m_a + m_b )}{m_b \mu} = \frac {g (m_a +m_b)}{\mu}

[/tex]

Plugging in the given values yields

[tex]

F_a = \frac {g (100 kg + 10 kg)}{.55} = 1960 newtons

[/tex]

However the answer given in the book is 172 newtons. I went back over my work and tried to get the given answer, but failed. If anyone has any idea on how to get 172 newtons, I'm more than welcome to them. All help is greatly appreciated.

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# Homework Help: Accelerating Masses

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