Froce Required to Overcome Friction of Two Boxes

[BraedenP]

Homework Statement

Two blocks, stacked one on top of the other, can move without friction on the horizontal surface shown in the figure. The surface between the two blocks is rough, however, with a coefficient of static friction equal to 0.44.

If a horizontal force F is applied to the 5.0kg bottom block, what is the maximum value F can have before the 2.0kg top block begins to slip?

Homework Equations

$$F=ma$$
$$F_s=\mu N$$

The Attempt at a Solution

Basically, I was trying to find the force on the bottom block that would make the force on the top block equal the force of static friction between the boxes.

So I calculated the normal force on the top block to be 19.62N. Then I multiplied that by $$\mu$$ to get $$F_s=8.6328N$$.

The acceleration in the top block that such a force would create is $$\frac{f}{m_1}=4.3164N$$. The force on the bottom block required to accelerate the system by this amount is $$m_2 \cdot a=21.582N$$

This answer, however, is wrong. I know I took a convoluted route to finding the answer, and obviously it didn't work... How would I approach a problem like this?

Thanks!

 

[quenderin]

You're almost there. In your last step you used Newton's second law to find a force on m_2. But this force isn't actually the horizontal force F. What is it?

 

[BraedenP]

You're almost there. In your last step you used Newton's second law to find a force on m_2. But this force isn't actually the horizontal force F. What is it?

Oh.. I need to consider the mass of the entire system when calculating the force, right?

So instead, the last equation would be:

$$(m_1+m_2) \cdot a=30.2148N$$

 

[quenderin]

yeah. alternatively, you should realize that the number 21.582N corresponds to the NET force on m_2. To find the horizontal force F, you've got to add the frictional force exerted by block 2 on block 1, which is 8.6328. You should get the same answer.

 

[BraedenP]

Oh, okay. That makes sense.

Thanks!