Calculate Force of Inclined Surface: 0.156kg at 55°

AI Thread Summary
To calculate the force on an inclined surface, a block with a mass of 0.156 kg moves at an angle of 55 degrees. The x and y components of the force are calculated as 1.25 and 0.877, respectively, with the normal force (N) being 0.877 N. The discussion highlights the need to sum forces in both x and y directions using the equations ΣFx: F(N)*sin(Θ)=0 and ΣFy: F(N)*cos(Θ)-mg=0. Clarification is sought on the term f(N) and the importance of considering the direction of sine and cosine in calculations. A diagram is recommended for better understanding of the forces involved.
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This is just a quick dumb question about a lab i had to do.. it involved putting a block on a piece of wood, and lifting the wood until the block moved. then we had to record the weight of the block and measure the angle that it took for the block to move...

for a block that had a mass of 0.156kg, it moved at 55 degrees.
i calculated the x component to be 1.25, and the y componenet to be 0.877. the force N(N) to be 0.877... but then it asks for f(N).. what is f(N)? how should i approach it.
 
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You should take into account the direction of the sin and cos and make a diagram.
Sum the forces is the x and y directions:
ΣFx: F(N)*sin(Θ)=+0
ΣFy: F(N)*cos(Θ)-mg=+0
Hope this helps. Sorry for the edits.
 
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