Question

A block of mass m/ = 4.0 kg is put on top of a block of mass m2 = 5.0 kg. To cause the top block to slip on the bottom one while the bottom one is held fixed, a horizontal force of at least 12 N must be applied to the top block. The blocks assembly is now placed on a horizontal, frictionless table, as shown in FIGURE 1. Find the magnitudes ofa.the maximum horizontal force that can be applied to the lower block so that the blockswill move togetherb.the resulting acceleration of the blocks.

Answer 1

If the horizontal force needed to move the top block is at least 12 N, we can calculate the static coefficient of friction with the formula below:

`\begin{gathered} F_(friction)=F_(normal)\cdot\mu_s\\ \\ 12=m\cdot g\cdot\mu_s\\ \\ 12=4\cdot10\cdot\mu_s\\ \\ \mu_s=(12)/(40)\\ \\ \mu_s=0.3 \end{gathered}`

(a)

Let's draw the force diagram of the system:

Using the second law of Newton for the complete system (considering the two blocks as one big block), we have:

`\begin{gathered} F=m\cdot a\\ \\ F_(applied)=(4+5)\cdot a\\ \\ a=(F_(applied))/(9) \end{gathered}`

Now, for the top block, we have:

`\begin{gathered} F=m\cdot a\\ \\ F_(applied)-F_(friction)=4\cdot a\\ \\ F_(applied)-12=(4)/(9)F_(applied)\\ \\ (5)/(9)F_(applied)=12\\ \\ F_(applied)=12\cdot(9)/(5)\\ \\ F_(applied)=21.6\text{ N} \end{gathered}`

(b)

The resulting acceleration will be: