# Blocks on an Incline

think about it, if the normal force = mgcos(theta) how can Fgx = mgcos(theta)?

i thought the fg would break up into fg in and fg out and the fg in would match up fn. Im looking at a past problem we've done but we solved for acceleration

ok so think about it like a triangle, the force collinear to the plane is Fgx, and the force orthogonal is Fgy, and Fn = Fgy.

okay so fgy = 9.8mcos(theta) and fgx = 9.8msintheta)

correct, ok so now that you know that, sum up all the forces in the x direction again.

the forces in the x direction are fgx and fst

9.8msin(theta) + (9.8m)cos(theta) * .7

fgx and fst are opposing forces, when you say "sum up the forces" you don't add the magnitudes. You have to set one direction to be positive and the other negative, so when they oppose forces one is negatively acting on the object and the other is positively acting on the object.

this is sooo confusing!!! thank you soooooo much for helping me !!!!

so would it be 9.8msin(theta) - (9.8m)cos(theta) * .7

yes it would be, now that you know what the sum of forces is recall what they are equal to from post 19-20.

9.8msin(theta) - (9.8m)cos(theta) * .7 = 0

9.8msin(theta) = (9.8m)cos(theta) * .7

9.8msin(theta) /(9.8m)cos(theta)= .7

sin(theta)/cos(theta) =.7

tan(theta) = .7

yes sir, and now you have the answer. ;)

thank you sooo much!!!!!!!!!!!!!!!!!! you are the best ever!!!!!!

No problem.