Ok so I already know how to solve it by looking at an example my teacher did in class... This is what I did... 4. A 2.00 kg block is held in equilibrium on an incline of angle θ = 70° by a horizontal force vector F applied in the direction shown in the figure below. If the coefficient of static friction between block and incline is μs = 0.300, determine the following. (a) the minimum value of F Step 1) Draw Free Body Diagram Step 2) Determine forces in x and y direction Fx fs F cos 70 -mg sin 70 Fy n -F sin 70 -mg cos 70 Step 3) Sum of x-forces and sum of y-forces/Equilibrium equations ƩFx = fs + F cos 70 + (-mg sin 70) = 0 ƩFy = n + (-F sin 70) + (-mg cos 70) = 0 ƩFy = n + (-F sin 70) + (-mg cos 70) = 0 ƩFy = n - F sin 70 - mg cos 70 = 0 ƩFy = n = F sin 70 + mg cos 70 ƩFy = n = 0.940F + (2.00)(9.8)cos 70 ƩFy = n = 0.940F + 6.704 Step 4) Solve for static friction fs fs = μsn fs = (0.300)n fs = (0.300)(0.940F + 6.704) fs = 0.282F + 2.0112 Step 5) Find the HORIZONTAL force vector by plugging in fs in ƩFx ƩFx = fs + F cos 70 + (-mg sin 70) = 0 ƩFx = (0.282F + 2.0112) + 0.342F - 18.418 = 0 ƩFx = 0.282F + 0.342F + 2.0112 - 18.418 = 0 ƩFx = 0.624F - 16.4068 = 0 ƩFx = 0.624F = 16.4068 ƩFx = F = 26.29 N (b) the normal force exerted by the incline on the block n = 0.940F + 6.704 n = 0.940(26.29) + 6.704 n = 31.4 N MY QUESTION IS... In STEP 2, why is the force "-mg sin 70" included in the X-DIRECTION and why is the force -mg cos 70 included in the Y-DIRECTION???? I thought the ONLY forces acting in the x-direction were static friction, fs and Fcos 70, while the only forces acting in the y direction were normal force, n, and -Fsin 70??? Help!