Calculate Force of Inclined Surface: 0.156kg at 55°

[xmflea]

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.

 

[Philosophaie]

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.

 

[baba89]

Hello,

Thank you for reaching out with your question about calculating the force of an inclined surface. To answer your question, the force (f) in this scenario refers to the net force acting on the block, which is the sum of the x and y components of the force. The x component (1.25) represents the force acting parallel to the surface, while the y component (0.877) represents the force acting perpendicular to the surface.

To calculate the net force (f), we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (f) is equal to the sum of the squares of the other two sides (x and y components). So, in this case, we can calculate the net force (f) as follows:

f = √(1.25^2 + 0.877^2) = √(1.5625 + 0.770129) = √2.332629 = 1.527 N

Therefore, the force (f) acting on the block is approximately 1.527 Newtons (N). I hope this helps clarify your question. Let me know if you have any further questions. Keep up the good work in your lab experiments!

Best,