When F = 2 , N,the frictional force between t | vedclass

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(B) Let $m_1 = 10 \, kg$ (upper block) and $m_2 = 5 \, kg$ (lower block).The coefficient of friction between the blocks is $\mu_1 = 0.1$ and between t

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$2$

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(B) Let $m_1 = 10 \, kg$ (upper block) and $m_2 = 5 \, kg$ (lower block).The coefficient of friction between the blocks is $\mu_1 = 0.1$ and between the lower block and the ground is $\mu_2 = 0.3$.The maximum static friction between the blocks is $f_{1 \max} = \mu_1 m_1 g = 0.1 	imes 10 	imes 10 = 10 \, N$.The maximum static friction between the lower block and the ground is $f_{2 \max} = \mu_2 (m_1 + m_2) g = 0.3 	imes (10 + 5) 	imes 10 = 0.3 	imes 150 = 45 \, N$.When an external force $F = 2 \, N$ is applied to the upper block,the force trying to move the system is $F = 2 \, N$.Since $F Consequently,the entire system $(15 \, kg)$ remains at rest because the applied force $F = 2 \, N$ is also less than $f_{2 \max} = 45 \, N$.Since the system is in equilibrium,the frictional force between the $5 \, kg$ block and the ground must balance the applied force $F$ to keep the system stationary.Therefore,the frictional force between the $5 \, kg$ block and the ground is $2 \, N$.

When $F = 2 \, N$,the frictional force between the $5 \, kg$ block and the ground is $.......... \, N$.

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