Calculate Force of Inclined Surface: 0.156kg at 55°

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SUMMARY

The discussion focuses on calculating the forces acting on a block with a mass of 0.156 kg placed on an inclined surface at an angle of 55 degrees. The user successfully calculated the x component of the force as 1.25 N and the y component as 0.877 N. The normal force, denoted as F(N), is required to solve the equations of motion, specifically using the equations ΣFx: F(N)*sin(Θ)=0 and ΣFy: F(N)*cos(Θ)-mg=0. A diagram illustrating the forces and their directions is recommended for clarity.

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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.
 
Last edited:

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