# Homework Help: Accelerating Masses

1. Jul 3, 2008

### Piamedes

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

$$\sum F = ma$$

$$F_f = \mu F_n$$

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.

$$\sum F_a = m_a a$$

$$m_a a = F_a - F_c$$

$$\sum F_b = m_b a$$

$$m_b a = F_c$$

Therefore

$$F_a = a ( m_a + m_b )$$

Then I substituted in for acceleration

$$a = \frac {F_c}{m_b}$$

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

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

$$\sum F_b = 0$$

$$0 = F_f - m_b g$$

$$m_b g = \mu F_c$$

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

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

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

Plugging in the given values yields

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

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.

2. Jul 4, 2008

### CompuChip

Can you be a little more clear about the things that you define? For example, you write
$$\sum F_a = m_a a$$ followed a little later by $$F_a = a ( m_a + m_b )$$. What is $F_a$ and what are you summing over? Is there a picture included in the question? Did you try drawing one with all the forces identified?

3. Jul 4, 2008

### Piamedes

Sorry about the lack of clarity.

$$\sum F_a = m_a a$$ means the sum of forces on block a

$$\sum F_b = m_b a$$ means the sum of forces on block b

$$F_a$$ means the applied force for which you are solving

$$F_f$$ means the frictional force between the two blocks

$$F_c$$ means the contact force between the two blocks

Diagram of the question

http://img78.imageshack.us/img78/329/diagramzk1.jpg [Broken]
http://g.imageshack.us/g.php?h=78&i=diagramzk1.jpg [Broken]

Free Body Diagrams

http://img357.imageshack.us/img357/4472/fbdtj6.jpg [Broken]
http://g.imageshack.us/g.php?h=357&i=fbdtj6.jpg [Broken]

I'm pretty sure about all the labeled forces except for the reaction to friction. However I can't see how that would affect the horizontal force and change the necessary applied force since there is no friction between the floor and block A. Thanks for the help.

Last edited by a moderator: May 3, 2017