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Block $A$ of mass $100 \mathrm{~kg}$ is kept above another block $B$ of mass $300 \mathrm{~kg}$. $A$ is tied to the wall $C$ with a horizontal string. The coefficient of friction between $A$ and $B$ is 0.35 and that between $B$ and the horizontal surface is 0.5 . Find the horizontal force $p$, necessary to move the block $B$.
Options:
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Verified Answer
The correct answer is:
$2350 \mathrm{~N}$
Given, the coefficient of friction between $A$ and $B, \mu_1=0.35$
The coefficient of friction between $B$ and surface, $\mu_2=0.5$
Mass of block $A, m_a=100 \mathrm{~kg}$
Mass of block $B, m_b=300 \mathrm{~kg}$
Frictional force between blocks $A$ and $B$,
$$
\begin{aligned}
f_1 & =\mu_1 m_a g \\
& =0.35 \times 100 \times 9.8 \\
& =0.35 \times 980 \\
& =343 \mathrm{~N}
\end{aligned}
$$
Frictional force between block $B$ and surface,
$$
\begin{aligned}
f_2 & =\mu_2\left(m_a+m_b\right) g \\
& =0.5(100+300) 9.8 \\
& =0.5 \times 4 \times 980=1960 \mathrm{~N}
\end{aligned}
$$
Minimum force required to move block $B$,
$$
\begin{array}{rlrl}
& & F & =f_1+f_2=343+1960 \\
\Rightarrow & & =2303 \mathrm{~N}
\end{array}
$$
Which is nearest to $2350 \mathrm{~N}$. Hence, option (c) is correct.
The coefficient of friction between $B$ and surface, $\mu_2=0.5$
Mass of block $A, m_a=100 \mathrm{~kg}$
Mass of block $B, m_b=300 \mathrm{~kg}$
Frictional force between blocks $A$ and $B$,
$$
\begin{aligned}
f_1 & =\mu_1 m_a g \\
& =0.35 \times 100 \times 9.8 \\
& =0.35 \times 980 \\
& =343 \mathrm{~N}
\end{aligned}
$$
Frictional force between block $B$ and surface,
$$
\begin{aligned}
f_2 & =\mu_2\left(m_a+m_b\right) g \\
& =0.5(100+300) 9.8 \\
& =0.5 \times 4 \times 980=1960 \mathrm{~N}
\end{aligned}
$$
Minimum force required to move block $B$,
$$
\begin{array}{rlrl}
& & F & =f_1+f_2=343+1960 \\
\Rightarrow & & =2303 \mathrm{~N}
\end{array}
$$
Which is nearest to $2350 \mathrm{~N}$. Hence, option (c) is correct.
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