# Two stacked blocks on a table. Friction problem

1. Nov 30, 2004

### HeRo

I need help with the following problem, I have tried a few different things, but none of them seem to be working:

Two blocks are stacked on a table. The upper block has a mass of 8.2 kg and the lower block has a mass of 35 kg. The coefficient of kinetic friction between the lower block and the table is 0.22 and the coefficient of static friction between the upper and lower blocks is 0.14. How much force can be applied to the lower block without having the upper block slide off?

Can someone explain how I would do this? It's got something to do with acceleration.

2. Nov 30, 2004

### Sirus

Like what? Please show us some attempt at solving the problem, or at least reasoning through it to some degree.

Last edited: Nov 30, 2004
3. Nov 30, 2004

### jdstokes

Apply Newton's second law to each block. Then solve simultaneous equations until you have an expression for F in terms of known variables. For the upper block
\begin{align*} \Sigma F_y & = 0 \\ n_1 - m_1g & = 0 \\ n_1 & = m_1g \\ \Sigma F_x & = m_1a \\ f_\mathrm{s} & = m_1a \\ f_\mathrm{s} & \leq \mu_\mathrm{s}n_1\\ m_1a & \leq \mu_\mathrm{s}m_1g \end{align*}
For the lower block
\begin{align*} \Sigma F_y & = 0 \\ n_2 - m_2g - n_1 & = 0 \\ n_2 & = n_1 + m_2g \\ n_2 & = g(m_1 + m_2) \\ \Sigma F_x & = m_2 a \\ F - f_\mathrm{k} - f_\mathrm{s} & = m_2a \\ F - \mu_\mathrm{k}n_2 - f_\mathrm{s} & = m_2a \\ F - \mu_\mathrm{k}g(m_1 + m_2) - f_\mathrm{s} & = \frac{m_2f_\mathrm{s}}{m_1} \\ F - \mu_\mathrm{k}g(m_1 + m_2) & = f_\mathrm{s}\left(1 +\frac{m_2}{m_1}\right) \\ f_\mathrm{s} & = \frac{F - \mu_\mathrm{k}g(m_1 + m_2)}{1 +\frac{m_2}{m_1}} \\ \frac{F - \mu_\mathrm{k}g(m_1 + m_2)}{1 +\frac{m_2}{m_1}}& \leq \mu_\mathrm{s}m_1g \\ \end{align*}
so
\begin{align*} F & \leq \mu_\mathrm{s}m_1g\left(1 +\frac{m_2}{m_1}\right) + \mu_\mathrm{k}g(m_1 + m_2) \end{align*}