Solving Newton's Laws: Find Friction Force on Block A

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Block A weighs 65.9N and has a static friction coefficient of 0.27, resulting in a maximum static friction force of approximately 17.793N. The system is in equilibrium, with a weight of 12.2N hanging from a string connected to Block A. The tension in the string exceeds the frictional force, indicating that all available friction is being utilized to keep Block A at rest. The calculated friction force on Block A is 18N, considering significant figures. The discussion emphasizes the relationship between tension and friction in maintaining equilibrium.
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Homework Statement



Block A in the figure weighs 65.9N. The coefficient of static friction between the block and the surface on which it rests is 0.27. The weight w is 12.2N and the system is in equilibrium.

block A is placed on a table, it is connected by a string to a wall block w is hung off that string to form a 90degree angle between A and w and a 45degree angle between w and the wall.

A) Find the friction force exerted on block A.

Homework Equations



Fr = u Fn
F = ma

The Attempt at a Solution



i tried to find the net force using f = ma but there is no horizonal acceleration... so i am confused. :S help??
 
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There is no horizontal acceleration, but there is an acceleration due to gravity on block W.

First you need to calculate the static force of friction on block A:
F=UsFn
F=(0.27*65.9)
F=17.793

Then you need to determine if the tension on the string from block W has enough force to move block A:
F(b)=MA
F(b)=12.2(9.8)
F(b)=119.56

Therefore your answer is 18, keeping significant figures in mind.

Conceptual/Additional Notes:
Since the tension force on the block is more than that of the frictional force, all the force of friction will be applied. If there wasn't enough force to move the block then only the force needed to keep block A at rest would have been applied. (If it were possible to apply more frictional force, the block would move opposite to the tension force).
 

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