Block A w/ Friction: Find Force & Max Weight

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SUMMARY

The discussion focuses on calculating the friction force and maximum weight for Block A, which weighs 64.2N with a static friction coefficient of 0.30. The friction force exerted on Block A is determined using the formula Ff = μN, resulting in a value of 19.26N. Additionally, the maximum weight that allows the system to remain in equilibrium is calculated as 52.9N, derived from the normal force equation where the normal force equals the weight of Block A minus the downward force of 11.3N.

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This discussion is beneficial for physics students, educators, and anyone studying mechanics, particularly those focusing on forces, friction, and equilibrium in systems.

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Homework Statement


Block A in the figure weighs 64.2N . The coefficient of static friction between the block and the surface on which it rests is 0.30. The weight is 11.3N and the system is in equilibrium.

The picture is a block A, sitting on a table. A string is attached horizontally and then the string breaks off into two separate strings. one connects to the wall with a angle of 45 degrees upward. and the other goes straight down with a weight on the end of it that weighs 11.3 N.

a)Find the friction force exerted on block A.
b)Find the maximum weight for which the system will remain in equilibrium.
 
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Homework Equations Ff=μNThe Attempt at a Solution a)Ff=μN Ff=.30(64.2N) Ff=19.26N b)The maximum weight for which the system will remain in equilibrium is the magnitude of the normal force (N) exerted on block A. The normal force is equal to the weight of the block (64.2N) minus the downward force of the 11.3N weight (i.e., 64.2-11.3=52.9N). Therefore, the maximum weight for which the system will remain in equilibrium is 52.9N.
 

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