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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?

\begin{figure*}
\epsfysize =1.5in
\centerline{\epsffile{e54.eps}}
\end{figure*}

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

\begin{displaymath}
a = \frac{R}{m_B},
\end{displaymath}

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

\begin{displaymath}
\mu R > m_B g,
\end{displaymath}

or

\begin{displaymath}
a > \frac{g}{\mu}.
\end{displaymath}

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

\begin{displaymath}
a = \frac{F}{m_A+m_B},
\end{displaymath}

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

\begin{displaymath}
F > \frac{(m_A+m_B) g}{\mu}.
\end{displaymath}

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

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


next up previous
Next: Conservation of energy Up: Newton's laws of motion Previous: Worked example 4.3: Raising
Richard Fitzpatrick 2006-02-02