Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
13:11 Sun 1 Dec ↑ DOF= 1 Tt ED Freeform Q2 a) Sammery of Ms = 0.7 knowns P=0 g=981 MK = 0.5 m= 50 kg Distances from ☐☐☐☐ Boxed Cente of gravity (6) rear wheels = 200 mm front wheels = 500mm Distence of G Summary of NB 屋 NC A = √900² + 1200² from the ground = 215 mm mm forces acting on object: Weight: mg = 9.81 (50) = 490.5 N normal forces: NB = ? Nc = .? Friction: No Mx (rear wheels only as Sam front wheels free to rotate) forces in equilibrium: ΣFy = 0 (as no movement in y 1 = NB + Nc - 490.5 Nc = 490.5-NB (i) = - P(0.4 -0.215) + Nc (0.5) - №8 (0.2) + F (0.215) = P=0 = 0.5 Nc - 0.2 NB - FF (0.215) = 0.5 NC - 0.2 NB - (NB. 0.5) (0.215) = (ii) (i) into (ii) = 0.5 (490.5-NB) - 0.2 NB - 0.1075 NB = 0 245.25-0.5NB - 0.2 NB - 0.1075 NB = 0 = -NB (0.1075 +0.5 +0.2) = - 245.25 NB = 303.7152 N Ne into (i) - Ne 490.5-303.7152 = 186.7848 N Fx= mao = P-F(0.215)=50a F= 0.5(303.7152) a = 151.8576 50 = = 151.8576 N -3.037152 2-3.04 m/s² .: The mower is decellerating at a rate of 23.04 m/s² J ล 76% Question 2: (a). The rear-wheel-drive lawn mower, when placed into gear while at rest, is observed to momentarily spin its rear tires as it accelerates. If the coefficients of friction between the rear tires and the ground are μs = 0.7 and μk = 0.5. Determine the forward acceleration 'a' of the mower. The mass of the mower and attached bag is 50 kg with centre of mass at G. Assume that the operator does not push on the handle, so that P=0. P 900 mm wwwwwwwwwwww.www. W215 mm 1000 mm B 500 C mm 200 mm Figure 2 (b). Define and explain the difference between compound gear train and planetary gear train. Explain your answer with three examples of each type. Draw sketches where requires.

13:11 Sun 1 Dec ↑ DOF= 1 Tt ED Freeform Q2 a) Sammery of Ms = 0.7 knowns P=0 g=981 MK = 0.5 m= 50 kg Distances from ☐☐☐☐ Boxed Cente of gravity (6) rear wheels = 200 mm front wheels = 500mm Distence of G Summary of NB 屋 NC A = √900² + 1200² from the ground = 215 mm mm forces acting on object: Weight: mg = 9.81 (50) = 490.5 N normal forces: NB = ? Nc = .? Friction: No Mx (rear wheels only as Sam front wheels free to rotate) forces in equilibrium: ΣFy = 0 (as no movement in y 1 = NB + Nc - 490.5 Nc = 490.5-NB (i) = - P(0.4 -0.215) + Nc (0.5) - №8 (0.2) + F (0.215) = P=0 = 0.5 Nc - 0.2 NB - FF (0.215) = 0.5 NC - 0.2 NB - (NB. 0.5) (0.215) = (ii) (i) into (ii) = 0.5 (490.5-NB) - 0.2 NB - 0.1075 NB = 0 245.25-0.5NB - 0.2 NB - 0.1075 NB = 0 = -NB (0.1075 +0.5 +0.2) = - 245.25 NB = 303.7152 N Ne into (i) - Ne 490.5-303.7152 = 186.7848 N Fx= mao = P-F(0.215)=50a F= 0.5(303.7152) a = 151.8576 50 = = 151.8576 N -3.037152 2-3.04 m/s² .: The mower is decellerating at a rate of 23.04 m/s² J ล 76% Question 2: (a). The rear-wheel-drive lawn mower, when placed into gear while at rest, is observed to momentarily spin its rear tires as it accelerates. If the coefficients of friction between the rear tires and the ground are μs = 0.7 and μk = 0.5. Determine the forward acceleration 'a' of the mower. The mass of the mower and attached bag is 50 kg with centre of mass at G. Assume that the operator does not push on the handle, so that P=0. P 900 mm wwwwwwwwwwww.www. W215 mm 1000 mm B 500 C mm 200 mm Figure 2 (b). Define and explain the difference between compound gear train and planetary gear train. Explain your answer with three examples of each type. Draw sketches where requires.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter8: Applications Of Trigonometry
Section: Chapter Questions
Problem 7RE
See similar textbooks
icon
Related questions
Question

may you please check my answer, and please dont use AI! may thanks, sam 

13:11 Sun 1 Dec ↑ DOF= 1 Tt ED Freeform Q2 a) Sammery of Ms = 0.7 knowns P=0 g=981 MK = 0.5 m= 50 kg Distances from ☐☐☐☐ Boxed Cente of gravity (6) rear wheels = 200 mm front wheels = 500mm Distence of G Summary of NB 屋 NC A = √900² + 1200² from the ground = 215 mm mm forces acting on object: Weight: mg = 9.81 (50) = 490.5 N normal forces: NB = ? Nc = .? Friction: No Mx (rear wheels only as Sam front wheels free to rotate) forces in equilibrium: ΣFy = 0 (as no movement in y 1 = NB + Nc - 490.5 Nc = 490.5-NB (i) = - P(0.4 -0.215) + Nc (0.5) - №8 (0.2) + F (0.215) = P=0 = 0.5 Nc - 0.2 NB - FF (0.215) = 0.5 NC - 0.2 NB - (NB. 0.5) (0.215) = (ii) (i) into (ii) = 0.5 (490.5-NB) - 0.2 NB - 0.1075 NB = 0 245.25-0.5NB - 0.2 NB - 0.1075 NB = 0 = -NB (0.1075 +0.5 +0.2) = - 245.25 NB = 303.7152 N Ne into (i) - Ne 490.5-303.7152 = 186.7848 N Fx= mao = P-F(0.215)=50a F= 0.5(303.7152) a = 151.8576 50 = = 151.8576 N -3.037152 2-3.04 m/s² .: The mower is decellerating at a rate of 23.04 m/s² J ล 76%
Transcribed Image Text:13:11 Sun 1 Dec ↑ DOF= 1 Tt ED Freeform Q2 a) Sammery of Ms = 0.7 knowns P=0 g=981 MK = 0.5 m= 50 kg Distances from ☐☐☐☐ Boxed Cente of gravity (6) rear wheels = 200 mm front wheels = 500mm Distence of G Summary of NB 屋 NC A = √900² + 1200² from the ground = 215 mm mm forces acting on object: Weight: mg = 9.81 (50) = 490.5 N normal forces: NB = ? Nc = .? Friction: No Mx (rear wheels only as Sam front wheels free to rotate) forces in equilibrium: ΣFy = 0 (as no movement in y 1 = NB + Nc - 490.5 Nc = 490.5-NB (i) = - P(0.4 -0.215) + Nc (0.5) - №8 (0.2) + F (0.215) = P=0 = 0.5 Nc - 0.2 NB - FF (0.215) = 0.5 NC - 0.2 NB - (NB. 0.5) (0.215) = (ii) (i) into (ii) = 0.5 (490.5-NB) - 0.2 NB - 0.1075 NB = 0 245.25-0.5NB - 0.2 NB - 0.1075 NB = 0 = -NB (0.1075 +0.5 +0.2) = - 245.25 NB = 303.7152 N Ne into (i) - Ne 490.5-303.7152 = 186.7848 N Fx= mao = P-F(0.215)=50a F= 0.5(303.7152) a = 151.8576 50 = = 151.8576 N -3.037152 2-3.04 m/s² .: The mower is decellerating at a rate of 23.04 m/s² J ล 76%
Question 2: (a). The rear-wheel-drive lawn mower, when placed into gear while at rest, is observed to momentarily spin its rear tires as it accelerates. If the coefficients of friction between the rear tires and the ground are μs = 0.7 and μk = 0.5. Determine the forward acceleration 'a' of the mower. The mass of the mower and attached bag is 50 kg with centre of mass at G. Assume that the operator does not push on the handle, so that P=0. P 900 mm wwwwwwwwwwww.www. W215 mm 1000 mm B 500 C mm 200 mm Figure 2 (b). Define and explain the difference between compound gear train and planetary gear train. Explain your answer with three examples of each type. Draw sketches where requires.
Transcribed Image Text:Question 2: (a). The rear-wheel-drive lawn mower, when placed into gear while at rest, is observed to momentarily spin its rear tires as it accelerates. If the coefficients of friction between the rear tires and the ground are μs = 0.7 and μk = 0.5. Determine the forward acceleration 'a' of the mower. The mass of the mower and attached bag is 50 kg with centre of mass at G. Assume that the operator does not push on the handle, so that P=0. P 900 mm wwwwwwwwwwww.www. W215 mm 1000 mm B 500 C mm 200 mm Figure 2 (b). Define and explain the difference between compound gear train and planetary gear train. Explain your answer with three examples of each type. Draw sketches where requires.
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
SEE MORE TEXTBOOKS