Forces at an Angle

1. Forces at an Angle

2. 30° FBD-H 1. NOT MOVING = ΣF = 0 FT FN Fg 2. What forces are acting on this block? 3. What is the Force due to gravity? Fg = w = mg w = 100 kg (10 m/s2) w = 1000 N

3. 30° FT FN 60° 30° w 4. Draw a FBD – skewed graph y FN x FT 30° w 5. Transpose the angles

4. 6. Resolve the x and y components of w y FN x FT wx wy 30° w SOHCAHTOA wy = ? wx = ? (A) cos Θ = A H 30° sin Θ = O H wy (H) w cos Θ = wy w sin Θ = wx w wx wy = w cos Θ wx = w sin Θ (O) wy = 1000 N cos 30° wx = 1000 N sin 30° wy = 866 N wx = 500 N

5. Memorize these:

6. 7. A simplified FBD helps you solve the rest. y FN x TO THE RIGHT IS POSITIVE FT TO THE LEFT IS NEGATIVE wx = 500N wy = 866N ΣFy=? FN = ? ΣFx=? FT = ? NOT MOVING = ΣFy = 0 NOT MOVING = ΣFx = 0 ΣFy= FN - wy (SIGNS ARE IMPORTANT) ΣFx= FT - wx 0 = FN - wy 0 = FT - wx FN = wy FT = wx FN = 866 N FT = 500 N

7. 40° FBD-I DO FOR HOMEWORK wy = ? wx = ? FN FN wy = w cos Θ FT FT sin Θ = O H wy = 1000 N cos 40° wx wy = 766 N sin Θ = wx w wy w w wx = w sin Θ wx = 1000 N sin 40° wx = 643 N

8. FN FT wx wy ΣFy=? FN = ? ΣFx=? FT = ? ΣFy = 0 ΣFx = 0 ΣFy= FN - wy ΣFx= FT - wx 0 = FN - wy 0 = FT - wx FN = wy FT = wx FN = 766 N FT = 643 N

9. 30° FBD-J 1. MOVING, NO FRICTION = ΣF = ma FN Fg 2. What forces are acting on this block? 3. What is the Force due to gravity? Fg = w = mg w = 100 kg (10 m/s2) w = 1000 N

10. 30° FN 60° 30° w 4. Draw a FBD – skewed graph FN y FN x w 30° w 5. Transpose the angles

11. 6. Resolve the x and y components of w y FN x wx wy 30° w SOHCAHTOA wy = ? wx = ? (A) cos Θ = A H 30° sin Θ = O H wy (H) w cos Θ = wy w sin Θ = wx w wx wy = w cos Θ wx = w sin Θ (O) wy = 1000 N cos 30° wx = 1000 N sin 30° wy = 866 N wx = 500 N

12. 7. A simplified FBD helps you solve the rest. y FN x TO THE RIGHT IS POSITIVE B/c it would move to the left and left is negative TO THE LEFT IS NEGATIVE wx = 500N wy = 866N ΣFy=? FN = ? ΣFx=? a = ? NOT MOVING up or down = ΣFy = 0 MOVING = ΣFx = -ma ΣFy= FN - wy ΣFx= - wx 0 = FN - wy -ma = - wx FN = wy a= wx -m FN = 866 N a= 500 N = 5 m/s2 100 kg

13. 8. Find vf =? 9. Find d =? vi = 0 d = vit + ½ at2 a = 5 m/s2 d = ½ at2 d = 5 m/s2 (10 s )2 2 t = 10 s vf = vi + at d = 250 m vf = at vf = 5 m/s2 (10 s) vf = 50 m/s

14. 30° FBD-K Now there’s friction. Predict a =? wy = ? wx = ? FN wy = w cos Θ Ff sin Θ = O H wy = 1000 N cos 30° wy = 866 N sin Θ = wx w w wx = w sin Θ wx = 1000 N sin 30° wx = 500 N

15. FN Ff wx wy ΣFy=? FN = ? Ff = ? ΣFx=? a = ? ΣFy = 0 Ff =µFN ΣFx = -ma ΣFy= FN - wy Ff =.1(866 N) ΣFx= Ff - wx 0 = FN - wy -ma = Ff - wx FN = wy a= Ff - wx -m FN = 866 N a= 86.6 N – 500 N= 4.13 m/s2 -100 kg

16. With and without friction: Find vf =? Find vf =? Find d =? Find d =? vi = 0 vi = 0 d = vit + ½ at2 d = vit + ½ at2 a = 5 m/s2 a = 4.13 m/s2 d = ½ at2 d = ½ at2 d = 5 m/s2 (10 s )2 2 d = 4.13 m/s2 (10 s )2 2 t = 10 s t = 10 s vf = vi + at vf = vi + at d = 250 m d = 206.5 m vf = at vf = at vf = 5 m/s2 (10 s) vf = 4.13 m/s2 (10 s) vf = 50 m/s vf = 41.3 m/s

17. Q = ? Looking for the threshold angle. What is the maximum angle that we can tilt the box before it starts to move? FBD-L FN Ff ΣFx = 0 w FN Ff SOHCAHTOA wx = ? wy = ? wx = w sin Θ wy = w cos Θ (A) Θ wy (H) w ΣFy = 0 w ΣFy= FN - wy wx 0 = FN - wy (O) FN = wy FN = w cos Θ

18. wx = w sin Θ ΣFx = 0 Ff = ? FN = w cos Θ ΣFx= Ff - wx Ff =µFN Ff= wx Ff =.4 FN FN Ff =.4 w cos Θ Ff= w sin Θ Ff .4 w cos Θ= w sin Θ wx wy

19. 20° FBD-M1 FN Ff w FN Ff SOHCAHTOA wy = ? wx = ? wy = w cos Θ (A) sin Θ = O H wy wy = 1000 N cos 20° (H) w w wy = 940 N sin Θ = wx w wx wx = w sin Θ (O) wx = 1000 N sin 20° wx = 342 N

20. FN Ff wx wy ΣFy=? FN = ? ΣFx=? Ff = ? µ = ? Ff =µFN ΣFy = 0 ΣFx = 0 µ = Ff FN ΣFy= FN - wy ΣFx= Ff - wx 0 = FN - wy 0 = Ff - wx µ = 342 N 940 N FN = wy Ff = wx FN = 940 N Ff = 342 N µ = .36