Block A of mass 3 kg rests on another block | vedclass

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(A) Let $m_A = 3 \ kg$ and $m_B = 7 \ kg$. The applied force $F = 50 \ N$ acts on block $B$.First, we check if the system moves. The maximum static fr

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(A) Let $m_A = 3 \ kg$ and $m_B = 7 \ kg$. The applied force $F = 50 \ N$ acts on block $B$.First, we check if the system moves. The maximum static friction force between block $B$ and the floor is $f_{max, floor} = \mu_{floor} (m_A + m_B) g$.$f_{max, floor} = 0.55 	imes (3 + 7) 	imes 10 = 0.55 	imes 100 = 55 \ N$.Since the applied force $F = 50 \ N$ is less than the maximum static friction force $f_{max, floor} = 55 \ N$, the system remains at rest.Because block $B$ does not move and there is no external force acting on block $A$ to cause it to slide relative to block $B$, the static friction force between $A$ and $B$ must be zero to maintain equilibrium.Therefore, the force of friction between $A$ and $B$ is $0 \ N$.

Similar Questions

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