Worked example 4.4: Suspended block

Question: Consider the diagram. The mass of block $A$ is $75 {\rm kg}$ and the mass of block $B$ is $15 {\rm kg}$. The coefficient of static friction between the two blocks is $\mu = 0.45$. The horizontal surface is frictionless. What minimum force $F$ must be exerted on block $A$ in order to prevent block $B$ from falling?

Answer: Suppose that block $A$ exerts a rightward force $R$ on block $B$. By Newton's third law, block $B$ exerts an equal and opposite force on block $A$. Applying Newton's second law of motion to the rightward acceleration $a$ of block $B$, we obtain

$$a = \frac{R}{m_B},$$

where $m_B$ is the mass of block $B$. The normal reaction at the interface between the two blocks is $R$. Hence, the maximum frictional force that block $A$ can exert on block $B$ is $\mu R$. In order to prevent block $B$ from falling, this maximum frictional force (which acts upwards) must exceed the downward acting weight, $m_B g$, of the block. Hence, we require

$$\mu R > m_B g,$$

or

$$a > \frac{g}{\mu}.$$

Applying Newton's second law to the rightward acceleration $a$ of both blocks (remembering that the equal and opposite forces exerted between the blocks cancel one another out), we obtain

$$a = \frac{F}{m_A+m_B},$$

where $m_A$ is the mass of block $A$. It follows that

$$F > \frac{(m_A+m_B) g}{\mu}.$$

Since $m_A=75 {\rm kg}$, $m_B = 15 {\rm kg}$, and $\mu = 0.45$, we have

$$F > \frac{(75+15)\times 9.81}{0.45} = 1.962\times 10^3 {\rm N}.$$