Calculating Frictional Force & Normal Force: Coeff of Friction

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The discussion revolves around calculating the normal force and frictional force acting on a block subjected to a downward force at an angle. For part (a), the normal force (Fn) is calculated as 24.6 N using the equation Fn = mg + 10sin30. In part (b), the frictional force was initially thought to be 19.7 N, but the correct calculation indicates it should be 8.7 N, considering the block's stationary state and the horizontal forces. The correct approach involves using the horizontal component of the applied force, leading to the conclusion that Fμ = 10 cos 30. This highlights the importance of analyzing forces accurately in static scenarios.
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Homework Statement


A force of 10N is applied downward at an angle of 30 deg with respect to the horizontal on a block. There is friction between the block and the floor and the block remains stationary. M=2kg
(a) What is the magnitude of the normal force acting on the block?

(b) The coefficient of friction between the floor and block is μ = 0.8. What is the magnitude of the frictional force acting on the block?

Homework Equations


Fn = mg + 10sin30
Fn(mu) = Fk

The Attempt at a Solution


I got part A -- Fn = mg + 10sin30 = 19.6 + 5.0 = 24.6

But for part B, I thought it would be 24.6*.8 = 19.7, but according to the answer key, it should be 8.7.
 
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The block is stationary on the floor so the sum of the horizontal forces must equal 0. So draw the free body diagram along with the horizontal forces acting on it.
 
So, Fμ = 10 cos 30 ?
 
Correct.
 
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