The blocks are in equilibrium. The friction f | vedclass

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(C) $1$. Analyze the forces on the $5 \, kg$ block: Since the system is in equilibrium,the tension $T$ in the string is equal to the weight of the $5

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$10 \, N$ up the plane

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(C) $1$. Analyze the forces on the $5 \, kg$ block: Since the system is in equilibrium,the tension $T$ in the string is equal to the weight of the $5 \, kg$ block. Thus,$T = m_2g = 5 	imes 10 = 50 \, N$.$2$. Analyze the forces on the $10 \, kg$ block along the inclined plane: The component of gravity acting down the plane is $m_1g \sin(37^\circ) = 10 	imes 10 	imes 0.6 = 60 \, N$.$3$. Determine the direction of friction: The gravitational force component $(60 \, N)$ is greater than the tension $(50 \, N)$ pulling the block up the plane. Therefore,the block has a tendency to slide down the plane. To maintain equilibrium,the friction force must act up the plane.$4$. Calculate the friction force: Let $f$ be the friction force. For equilibrium along the plane,$T + f = m_1g \sin(37^\circ)$. Substituting the values,$50 + f = 60$,which gives $f = 10 \, N$ up the plane.

The blocks are in equilibrium. The friction force acting on the $10 \, kg$ block is:

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