Newton's Laws Ramps Page 1a Schaum's 3

1. Newton's Laws Ramps Page 1a Schaum's 3 F + + ΣF = ma F - mg sin θ = 0 F = mg sin θ F = 200 sin 30 F = 100 N mg sin θ Fn mg cos θ Ff + ΣF = ma ΣF = ma + mg sin θ - Ff = ma mg sin θ - μFn = ma 20 (9.8) sin 30 - 0.30(169.74) = 20 a a = 2.35 m/s2 mg cos θ - Fn = 0 Fn = mg cos θ Fn = 20 (9.8) cos 30 Fn = 169.74 N mg sin θ Fn mg cos θ 3.26 A 200 N wagon is to be pulled up a 300 incline at constant speed. How large a force parallel to the incline is needed if friction effects are negligible? 3.27 A 20 kg box sits on a 300 incline. The coefficient of kinetic friction between box and incline is 0.30. Find the acceleration of the box down the ramp.

2. Newton's Laws Ramps Page 1b Schaum's 3 400 400 + 500 sin 40 Ff 500 400 500 cos 40 Fn mg sin θ mg cos θ ΣF = ma + + ΣF = ma mg cos θ + 500 sin 40 - Fn = 0 25 ( 9.8) cos 40 + 500 sin 40 - Fn = 0 187.68 + 321.39 - Fn = 0 Fn = 509.07 N 500 cos 40 - Ff - mg sin θ = ma 500 cos 40- μFn - mg sin θ = ma 383.02 - μ (509.07) - 25 (9.8) sin 40 = 25(.75) μ = 0.406 3.28 When a force of 500 N pushes on a 25 kg box as shown, the acceleration of the box up the incline is 0.75 m/s 2. Find the coefficient of kinetic friction between box and incline.

3. Newton's Laws Ramps Page 1b Schaum's 3 3.34 The two boxes shown have identical masses of 40 kg. Both experience a sliding friction force with μ = 0.15 . Find the acceleration and the tension of the system. + + A B 300 T + ΣF = ma Ff + + mg cos θ - Fn = 0 40 (9.8) cos 30= Fn Fn = 339.48 N Ff T mg cos θ mg sin θ Fn + mg Fn + ΣF = ma ΣF = ma mg sin θ - Ff - T = ma 40 (9.8) sin 30- μFn - T = 40a 196 - 0.15 (339.48) - T = 40a 196 - 50.92 - T = 40a 145.08 - T = 40a T - Ff = ma T - 0.15(40)(9.8) = 40a T - 58.8 = 40a 145.08 - T = 40a T - 58.8 = 40a T - 58.8 = 40a T - 58.8 = 40(1.08) T = 102N 86.28 = 80a a = 1.08 m/s2

4. mg sin θ Fn mg cos θ Newton's Laws Ramps Page 2a Schaum's 3 3.61 A 12 kg box is released from the top of an incline that is 5 m long and makes an angle of 400 to the horizontal. A 60 N friction force impedes the motion of the box. a) What will be the acceleration of the box b) how long will it take to reach the bottom of the incline? and c) What is the coefficient of friction between the box and incline? Ff + + ΣF = ma mg sin θ - Ff = ma 12(9.8) sin 40 - 60 = 12a a = 1.30 m/s2 x = V0t + 1/2 at2 5 = 0 + 1/2 (1.30)t2 t = 2.77 seconds Ff = μFn 60 = μmgcosθ 60 = μ(12)(9.8)cos40 μ = 0.666

5. + + F ΣF = ma F - mg sin θ = ma F - 15 (9.8) sin 30 = 15(1.2) F = 91.5 N mg sin θ Fn mg cos θ + F + ΣF = ma F + mg sin θ = ma F + 15 (9.8) sin 30 = 15(1.2) F = -55.5 N Need to push up so it doesn't accelerate greater than 1.2 m/s2 mg sin θ Fn mg cos θ Newton's Laws Ramps Page 2a Schaum's 3 3.63 An inclined plane makes an angle of 300 with the horizontal. Find the constant force applied parallel to the plane, required to cause a 15 kg box to slide a) up the plane with acceleration 1.2 m/s2 and b) down the incline with acceleration 1.2 m/s2. Neglect friction forces.

6. + F sin 30 Ff + ΣF = ma + ΣF = ma F F cos θ - mg sin θ - Ff = ma F cos 30 - 20(9.8) sin 30 - 80 = 20 (.75) F = 222.86 N F cos θ - mg sin θ - Ff = 0 F cos 30 - 20(9.8) sin 30 - 80 = 0 F = 205.54 N 300 F cos 30 Fn mg sin θ mg cos θ + ΣF = ma 200 sin 20 + Ff mg cos θ + 200 sin 20 - Fn = 0 15 ( 9.8) cos 20 + 200 sin 20 - Fn = 0 187.68 + 321.39 - Fn = 0 Fn = 206.53 N 200 N 200 Fn 200 cos 20 mg sin θ mg cos θ Newton's Laws Ramps Page 2b Schaum's 3 3.64 A horizontal force F is exerted on a 20 kg box to slide it up a 300 incline. The friction force retarding the motion is 80 N. How large must F be if the acceleration of the moving box is to be a) zero and b) 0.75 m/s2 ? 3.67 A horizontal force of 200 N is required to cause a 15 kg block to slide up a 200 incline with an acceleration of 25 cm/s2. Find a) the friction force on the block and b) the coefficient of friction. + ΣF = ma F cos θ - mg sin θ - Ff = ma 200 cos 20 - 15(9.8) sin 20 - Ff = 15 (.25) Ff = 133.91 N Ff = μFn 133.91 = μ(206.53) μ = 0.648

7. T + T mg sin θ + Fn mg cos θ W ΣF = ma + + ΣF = ma T - mg sin θ = ma T- 30(9.8) sin 25 = 30a T - 124.25 = 30a W - T = ma 20(9.8) - T = 20a 196 - T = 20a T + T Ff + Fn mg sin θ mg cos θ W + ΣF = ma + ΣF = ma T - mg sin θ - Ff = ma T- 30(9.8) sin 25 - .20(30)(9.8) cos 25 = 30a T - 177.54 = 30a W - T = ma 20(9.8) - T = 20a 196 - T = 20a mg sin θ Fn mg cos θ Newton's Laws Additional Page 3a Schaum's 3 3.65 An inclined plane making an angle of 250 with the horizontal has a pulley on its top. A 30 kg block on the plane is connected to a freely hanging 20 kg block by means of a cord passing over the pulley. Compute the distance the 20 kg block will fall in 2 s starting from rest. Neglect friction. a = 1.44 m/s2 y = 1/2 at2 y = 1/2 (1.44) (2)2 y = 2.88 m 3.66 Repeat problem 3.65 if the coefficient of friction between block and plane is 0.20. a = 0.37 m/s2 y = 1/2 at2 y = 1/2 (0.37) (2)2 y = 0.74m

8. 10 kg μ = 0.20 6 kg 9 kg T + TL + 10 TR Ff W mg Fn + + ΣF = 0 + ΣF = ma ΣF = ma mg - Fn = 0 10(9.8) - Fn = 0 Fn = 98 TL - W = ma TL -6(9.8) = 6a TL -58.8 = 6a TR - TL - Ff = ma TR - TL - .2(98) = 10a TR - TL - 19.6 = 10a Newton's Laws Additional Page 3b Schaum's 3 3.74 Find the acceleration of the system and the tension in the cord on the left and in the cord on the right. TL 9 + 6 W + ΣF = ma W - TR = ma 9(9.8) - TR = 9a 88.2 - TR = 9a a = 0.39 m/s2 TL = 61.14 N TR = 84.69 N