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Forces at an Angle

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

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30° w FNFN FTFT 4. Draw a FBD – skewed graph y xFNFN w FTFT 60°30° 5. Transpose the angles 30°

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y xFNFN w FTFT 6. Resolve the x and y components of w wywy wxwx w 30° wywy wxwx (H) (A) (O) SOHCAHTOA w x = ? sin Θ = O H sin Θ = w x w w x = w sin Θ w x = 1000 N sin 30° w x = 500 N w y = ? cos Θ = A H cos Θ = w y w w y = w cos Θ w y = 1000 N cos 30° w y = 866 N

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Memorize these: sincos 0°0°01 30° ° ° °10

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y x FTFT w y = 866N w x = 500N 7. A simplified FBD helps you solve the rest. FNFN ΣF y =? F N = ? NOT MOVING = ΣF y = 0 ΣF y = F N - w y (SIGNS ARE IMPORTANT) TO THE LEFT IS NEGATIVE TO THE RIGHT IS POSITIVE 0 = F N - w y F N = w y F N = 866 N ΣF x =? F T = ? NOT MOVING = ΣF x = 0 ΣF x = F T - w x 0 = F T - w x F T = w x F T = 500 N

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FBD-I 40° DO FOR HOMEWORK FNFN w FTFT FNFN w FTFT wywy wxwx w x = ? sin Θ = O H sin Θ = w x w w x = w sin Θ w x = 1000 N sin 40° w x = 643 N w y = ? w y = w cos Θ w y = 1000 N cos 40° w y = 766 N

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ΣF y =? F N = ? ΣF y = 0 ΣF y = F N - w y 0 = F N - w y F N = w y F N = 766 N ΣF x =? F T = ? ΣF x = 0 ΣF x = F T - w x 0 = F T - w x F T = w x F T = 643 N FNFN FTFT wywy wxwx

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

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30° 4. Draw a FBD – skewed graph y xFNFN w 60°30° 5. Transpose the angles 30° FNFN w FNFN w

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y xFNFN w 6. Resolve the x and y components of w wywy wxwx w 30° wywy wxwx (H) (A) (O) SOHCAHTOA w x = ? sin Θ = O H sin Θ = w x w w x = w sin Θ w x = 1000 N sin 30° w x = 500 N w y = ? cos Θ = A H cos Θ = w y w w y = w cos Θ w y = 1000 N cos 30° w y = 866 N

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y x w x = 500N 7. A simplified FBD helps you solve the rest. FNFN ΣF y =? F N = ? NOT MOVING up or down = ΣF y = 0 ΣF y = F N - w y TO THE LEFT IS NEGATIVE TO THE RIGHT IS POSITIVE 0 = F N - w y F N = w y F N = 866 N ΣF x =? a = ? MOVING = ΣF x = -ma ΣF x = - w x -ma = - w x a = w x -m a = 500 N = 5 m/s kg B/c it would move to the left and left is negative

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8. Find v f =? v i = 0 a = 5 m/s 2 t = 10 s v f = v i + at v f = at v f = 5 m/s 2 (10 s) v f = 50 m/s 9. Find d =? d = v i t + ½ at 2 d = ½ at 2 d = 5 m/s 2 (10 s ) 2 2 d = 250 m

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FBD-K 30° Now theres friction. Predict a =? FNFN w FfFf w x = ? sin Θ = O H sin Θ = w x w w x = w sin Θ w x = 1000 N sin 30° w x = 500 N w y = ? w y = w cos Θ w y = 1000 N cos 30° w y = 866 N

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FNFN FfFf wywy wxwx ΣF y =? F N = ? ΣF y = 0 ΣF y = F N - w y 0 = F N - w y F N = w y F N = 866 N ΣF x =? a = ? ΣF x = -ma ΣF x = F f - w x -ma = F f - w x a = F f - w x -m a = 86.6 N – 500 N= 4.13 m/s kg F f = ? F f = µF N F f =.1(866 N)

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With and without friction: Find v f =? v i = 0 a = 5 m/s 2 t = 10 s v f = v i + at v f = at v f = 5 m/s 2 (10 s) v f = 50 m/s Find v f =? v i = 0 a = 4.13 m/s 2 t = 10 s v f = v i + at v f = at v f = 4.13 m/s 2 (10 s) v f = 41.3 m/s Find d =? d = v i t + ½ at 2 d = ½ at 2 d = 5 m/s 2 (10 s ) 2 2 d = 250 m Find d =? d = v i t + ½ at 2 d = ½ at 2 d = 4.13 m/s 2 (10 s ) 2 2 d = m

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FBD-L = ? Looking for the threshold angle. What is the maximum angle that we can tilt the box before it starts to move? ΣF x = 0 w FNFN FfFf FNFN w FfFf w Θ wywy wxwx (H) (A) (O) SOHCAHTOA w x = ? w x = w sin Θ w y = ? w y = w cos Θ ΣF y = 0 ΣF y = F N - w y 0 = F N - w y F N = w y F N = w cos Θ

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F f = ? F f = µF N F f =.4 F N F f =.4 w cos Θ FNFN FfFf wywy wxwx ΣF x = 0 ΣF x = F f - w x F f = w x w x = w sin Θ F f = w sin Θ.4 w cos Θ = w sin Θ

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FBD-M1 20° w FNFN FfFf FNFN w FfFf w wywy wxwx (H) (A) (O) SOHCAHTOA w x = ? sin Θ = O H sin Θ = w x w w x = w sin Θ w x = 1000 N sin 20° w x = 342 N w y = ? w y = w cos Θ w y = 1000 N cos 20° w y = 940 N vivi NA vfvf a0

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ΣF y =? F N = ? ΣF y = 0 ΣF y = F N - w y 0 = F N - w y F N = w y F N = 940 N F f = µF N FNFN FfFf wywy wxwx ΣF x =? F f = ? ΣF x = 0 ΣF x = F f - w x 0 = F f - w x F f = w x F f = 342 N µ = ? µ = F f F N µ = 342 N 940 N µ =.36

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