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Understanding Newton's Laws & Friction Scenarios

Explore practical examples and calculations related to Newton's Laws and friction in various scenarios without ramps. Learn how forces, accelerations, and coefficients of friction are interconnected in physics problems.

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Understanding Newton's Laws & Friction Scenarios

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  1. Newton's Laws Friction (No Ramps) Page 1a Schaum's 3 + + + ΣF = ma Vf2 = Vo2 + 2ax 0 = .82 + 2(-1.75)x x = 0.18m = 18 cm Ff = μFn 0.7 = μ(.4)(9.8) μ = 0.179 Ff = ma -0.7 = .4a a = - 1.75 m/s2 W Fn + 3.12 A horizontal force of 140 N is needed to pull a 6 kg box across a horizontal floor at constant speed. What is the coefficient of friction between floor and box? + 140 = Ff 140 = μFn 140 = μ(6)(9.8) μ = 0.238 ΣF = 0 F- Ff = 0 F = Ff Ff 140N Fn W 140 = Ff 3.16 A 400 g block with an initial speed of 80 cm/s slides along a horizontal tabletop against a friction force of 0.70 N. a) How far will it slide before stopping? b) What is the coefficient of friction between the block and the tabletop? Ff 3.17 A 600 kg car is moving on a level road at 30 m/s. a) How large a retarding force is required to stop it in a distance of 70 m? b) What is the minimum coefficient of friction between roadway and tires if this is possible? + Ff = μFn 3858 = μ(600)(9.8) μ = 0.656 Vf2 = Vo2 + 2ax 0 = 302 + 2a(70) a = - 6.43 m/s2 ΣF = ma Ff = ma Ff = 600(6.43) Ff = 3858 N Ff = 3.86 KN Ff W Fn

  2. + 400N 500 Ff W Fn Newton's Laws Friction (No Ramps) Page 2a Schaum's 3 3.25 A force of 400 N pushes on a 25 kg box at an angle of 500 to the horizontal. Starting from rest, the box acheives a velocity of 2 m/s in a time of 4 s. Find the coefficient of kinetic friction between box and floor. + a = Vf - Vo t a = 2 - 0 4 a = 0.5 m/s2 ΣF = 0 + ΣF = ma 400cos 500 - Ff = 25(.5) Ff = 244.62N Ff = μFn 244.62 = μ551.42 μ = 0.444 400 sin 50 - Fn + 25(9.8) = 0 Fn = 551.42 N 3.33 The coefficient of static friction between a box and the flat bed of a truck is 0.60. What is the maximum acceleration the truck can have along level ground if the box is not to slide? + + μg = a 0.6(9.8) = a a = 5.88 m/s2 ΣF = ma Ff Ff = ma μFn = ma μmg = ma μg = a W Fn

  3. Newton's Laws Friction (No Ramps) Page 1b Schaum's 3 + + ΣF = ma 400 - Ff = ma 400 - 0.5(70)(9.8) = 70a a = 0.81 m/s2 + + ΣF = 0 ΣF = ma 400cos 300 - Ff = 70a 0.866 (400) - 0.5(486) = 70a a = 1.48 m/s2 400 sin θ + Fn - 70 (9.8) = 0 Fn = 486 N 3.22 A 70 kg box is slid along the floor by a 400 N force. The coefficient of friction between the box and floor is 0.50 when the box is sliding. Find the acceleration of the box. Ff 400 N W Fn 3.23 Suppose the 70 kg box in 3.22 is pulled by a 400 N force at an angle of 300 to the horizontal. The coefficient of friction is 0.50. Find acceleration. + 400 N Ff W Fn

  4. + T ΣF = ma A + Ff W - T = ma 15(9.8) - T = 15a T + + B W Fn ΣF = ma a = 2.45 m/s 2 y = V0 t + 1/2 at2 y = 1/2 (2.45)(42) y = 11.03 m W T - Ff = ma T - .2(25)(9.8) = 25a T - 49 = 25a 3.32 How large a horizontal force in addition to the tension must pull on A in 3.31 to give it an acceleration of 0.75 m/s2 toward the left. T μ = 0.20, ma = 25 kg and mb = 15 kg. ΣF = ma + + + T - W = ma T - 147 = 15(.75) T = 158.25 N ΣF = ma + F - T - Ff = ma F - T - .2(25)(9.8) = 25(.75) F - T - 49 = 18.75 T F Ff W W Fn F - 158.25 - 49 = 18.75 F = 226 N Newton's Laws Friction (No Ramps) Page 2b Schaum's 3 3.31 In the diagram, μ = 0.20, ma = 25 kg and mb = 15 kg. How far will block B drop in the first 3 seconds after the system is released?

  5. + + F + ΣF = 0 ΣF = ma + ΣF = ma 300 Fcos 300 - Ff = 20(.4) 0.866 F - 30 = 8 F = 43.88 N Fsin 30 + Fn - 20 (9.8) = 0 Fn = 196 - 0.5F Fcos 300 - Ff = 0 0.866 F - 30 = 0 F = 34.64 N 30N W Fn 2 kg F 1.5 kg 1 kg + + + T12 F T23 T23 T12 Ff Ff Ff Newton's Laws Friction (No Ramps) Page 3a Schaum's 3 W Fn W Fn W Fn + + + ΣF = ma ΣF = ma ΣF = ma F - T12 - Ff = ma F - T12 - .2(1.5)(9.8) = 1.5(3) F - T12 = 7.44 T23 - Ff = ma T23 - .2(1)(9.8) = 1(3) T23 = 4.96 N T12- T23 - Ff = ma T12- T23 - .2(2)(9.8) = 2(3) T12 - T23 = 9.92 T12 - T23 = 9.92 T12 - 4.96 = 9.92 T12 = 14.88 N F - T12 = 7.44 F - 14.88 = 7.44 F = 22.32 N T23 = 4.96 N 3.60 A 20 kg wagon is pulled along the level ground by a rope inclined at 300 above the horizontal. A friction force of 30 N opposes the motion. How large is the pulling force if the wagon is moving with a) constant speed b) an acceleration of 0.40 m/s2 ? 3.71 How large a force F is needed to give the blocks an acceleration of 3 m/s2? μ = 0.20. How large a force does the 1.5 kg block exert on the 2 kg block? 1 2 3

  6. 5 kg F = 30 N 4 kg Fig 3.68 + + + ΣF = ma + ΣF = ma F - T = ma 30 - T = 5a 30 - 4a = 5a a = 3.33 m/s2 T T 30N T = ma T = 4a W Fn W Fn T = 4a T = 4(3.33) T = 13.32 N + T = 4a + 11.76 T = 4(.39) + 11.76 T = 13.32 N + ΣF = ma + + ΣF = ma F - T - Ff = ma 30 - T - .3(5)(9.8) = 5a 30 - T - 14.7 = 5a 30 - (4a + 11.76) - 14.7 = 5a 9a = 3.54 a = 0.39 m/s2 Ff T - Ff = ma T - .3(4)(9.8) = 4a T - 11.76 = 4a T = 4a + 11.76 Ff T T 30N W Fn W Fn Newton's Laws Friction (No Ramps) Page 3b Schaum's 3 3.68 In Figure 3.68, friction forces are negligible. What is the acceleration of the blocks and the tension in the cord connecting them? 3.69 In figure 3.68, the coefficient of friction is 0.30. What is the acceleration of the blocks and the tension in the cord connecting them?

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