A block A of mass 4, kg is placed on another | vedclass

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(C) Let the coefficient of friction between blocks $A$ and $B$ be $\mu$. The maximum static friction force between $A$ and $B$ is $f_{max} = \mu m_A g

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) Let the coefficient of friction between blocks $A$ and $B$ be $\mu$. The maximum static friction force between $A$ and $B$ is $f_{max} = \mu m_A g = \mu 	imes 4 	imes 10 = 40\mu$.When a force $F_A = 12\, N$ is applied on $A$,the acceleration of the system is $a = \frac{F_A}{m_A + m_B} = \frac{12}{4 + 5} = \frac{12}{9} = \frac{4}{3}\, m/s^2$.For block $A$ to move with $B$,the friction force $f$ must provide the acceleration to $A$: $f = m_A a = 4 	imes \frac{4}{3} = \frac{16}{3}\, N$.Since this is the limiting case,$f = f_{max} \Rightarrow 40\mu = \frac{16}{3} \Rightarrow \mu = \frac{16}{120} = \frac{2}{15}$.Now,let a maximum force $F_B$ be applied to block $B$. The acceleration of the system is $a' = \frac{F_B}{m_A + m_B} = \frac{F_B}{9}$.For block $A$ to move with $B$,the friction force $f$ must provide the acceleration to $A$: $f = m_A a' = 4 	imes \frac{F_B}{9}$.Since $f \le f_{max}$,we have $4 	imes \frac{F_B}{9} \le 40 	imes \frac{2}{15}$.$F_B \le 40 	imes \frac{2}{15} 	imes \frac{9}{4} = 10 	imes 2 	imes \frac{3}{5} 	imes 3 = 36\, N$. Wait,re-calculating: $F_B \le 40 	imes \frac{2}{15} 	imes \frac{9}{4} = 10 	imes 2 	imes \frac{9}{15} = 20 	imes 0.6 = 12\, N$ is incorrect. Let's re-evaluate: $f_{max} = 40 	imes \frac{2}{15} = \frac{16}{3}$. $4 	imes \frac{F_B}{9} = \frac{16}{3} \Rightarrow F_B = \frac{16}{3} 	imes \frac{9}{4} = 4 	imes 3 = 12\, N$. Actually,the question implies $F_B$ is applied to $B$ only. $F_B - f = m_B a' \Rightarrow F_B - \frac{16}{3} = 5 	imes \frac{F_B}{9} \Rightarrow F_B(1 - 5/9) = 16/3 \Rightarrow F_B(4/9) = 16/3 \Rightarrow F_B = 16/3 	imes 9/4 = 12\, N$. Given the options,let's re-check the initial $\mu$ calculation. $12 - f = 4a$ and $f = 5a$. $12 = 9a \Rightarrow a = 4/3$. $f = 5(4/3) = 20/3$. $f_{max} = \mu(4)(10) = 40\mu = 20/3 \Rightarrow \mu = 1/6$. Now for $B$: $F_B - f = 5a'$ and $f = 4a'$. $F_B = 9a'$. Max $f = 40(1/6) = 20/3$. $4a' = 20/3 \Rightarrow a' = 5/3$. $F_B = 9(5/3) = 15\, N$.

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