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Say there are a block ($m_1=10\text{ kg}$) resting on top of a bracket ($m_2=5.0\text{ kg}$), the bracket sitting on a frictionless surface. See the drawing below. Given are the coefficients of static and kinetic friction: $\mu_s = 0.40$ and $\mu_k = 0.30$.

Drawing of the block and bracket.

The question is, what is the maximal force $F$ that can be applied without sliding the block on the bracket, and what is the corresponding acceleration $a$ of the bracket?

I know this question is similar to this question, but I don't understand why $f_s = m_2a$ for the bottom block.

What I did: $$f_{s,max} = - \mu_sF_n = -\mu_s m_1 g$$ $$F = -f_{s,max} = \mu_sm_1g$$ Also $F=m_{total}a = (m_1 + m_2)a$, so $$\mu_sm_1g = (m_1 + m_2)a$$ $$a = \frac{\mu_sm_1g}{m_1 + m_2}$$

Plugging in the numbers, I get $$a = \frac{0.40\times 10.0 \times 9.81}{10.0 + 5.0} = 2.616 \text{ m/s}^2$$

However, given answer is $a = 1.6 \text{ m/s}^2$.

What am I doing wrong and/or how do I correctly solve this problem?

Thanks in advance.

Community
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Kevin
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  • Your question seems to assume that the top block is being pushed, but if so the correct answer should be 1.96 m/s$^2$. If the bottom block is being pushed, the acceleration would be 7.85 m/s$^2$. – Suzu Hirose Oct 02 '16 at 06:55

1 Answers1

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Wherever the force $F$ (assumed horizontal) is applied the maximum static frictional force acting on the two blocks is $\mu_s m_1g$.

That maximum static frictional force will accelerate the top block or accelerate the bottom block (not both blocks at once as you have in your solution) and the frictional force will hinder the acceleration of the other block which will have the force $F$ acting on it.

Farcher
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  • I don't understand this answer. The point is that the two blocks are not in relative motion. Why did you write "or" in bold face like that? I think both blocks are supposed to move as one. – Suzu Hirose Oct 02 '16 at 00:51
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    They do move as one but one block has only a frictional force on it and the other block has a horizontal force $F$ on it and a frictional force in the opposite direction to force $F$. The force $F$ does accelerate both blocks. As force $F$ is increased the acceleration of both blocks increases but if force $F$ which is acting on one of the blocks gets too big the frictional force on the other block is not large enough to accelerate it at the same rate as the block with force $F$ acting on it. – Farcher Oct 02 '16 at 06:10
  • OK sorry I read the other answer to the other question and worked it out. – Suzu Hirose Oct 02 '16 at 06:55