Block A w/ Friction: Find Force & Max Weight

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The problem involves Block A weighing 64.2N, with a static friction coefficient of 0.30. The friction force exerted on Block A is calculated as 19.26N using the formula Ff=μN. To determine the maximum weight for equilibrium, the normal force is found by subtracting the downward force of 11.3N from Block A's weight, resulting in 52.9N. Thus, the maximum additional weight that can be added while maintaining equilibrium is 52.9N.
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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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