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Calculating Forces on Inclined Plans and Acceleration Problems

This chapter discusses various force diagrams and calculations for objects on inclined planes. It provides examples for calculating the applied force required to accelerate a block up a ramp at a specified angle, considering friction and other forces acting on the object. The text includes step-by-step solutions to problems involving mass, friction coefficients, and net forces. Problems also explore the relationship between angles, gravitational forces, and the required applied forces to achieve desired accelerations.

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Calculating Forces on Inclined Plans and Acceleration Problems

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  1. Chapter 4 Other Force Diagrams

  2. W|| W 28° Ff Find Ff : Find FNET : Find FA : FN FA =.675(51.9) = 6(3.5) FA – 27.6 – 35 = 21 = 35.0 N = 21.0 N 83.6 N But first one RAMP PROBLEM for review………. With what force must you push to accelerate a 6.00 kg block 3.50 m/s2 UP a 28° ramp with  = .675? FN = W FA – W|| - Ff = FNET 6 58.8 51.9 27.6 m = W = W= W||= Ff =  FN FNET = ma FA – W|| - Ff = FNET

  3. Handout

  4. What is the acceleration of a 20.0 kg object when pulled with a rope that makes an angle of 22° with the horizontal? The applied force on the rope is 250 N and  = .333. FA W FN F 250 F|| 250 sin = cos = 250 F Ff F|| F 250 F|| FN + F = W F|| - Ff = FNET m = W= FA= F= F||= 20 250 196 93.7 232

  5. What is the acceleration of a 20.0 kg object when pulled with a rope that makes an angle of 22° with the horizontal? The applied force on the rope is 250 N and  = .333. FA W FN Find FN: Find Ff : Find FNET: Find a: Ff F 250 F|| FN + F = W F|| - Ff = FNET m = W= FA= F= F||= 20 196 250 93.7 232 Ff =  FN FN + F = W F|| - Ff = FNET FNET = ma FN+ 93.7 = 196 =.333(102) 232 – 34 = FNET 198 = 20 a 102 N 34.0 N 198 N 9.90 m/s2

  6. What is the coefficient of friction if pulling with 55.0 N on a handle that makes an angle of 35° with the ground gives a 60.0 N object an acceleration of 3.00 m/s2? FN F 55 F|| Ff W Find FNet: Find Ff : Find FN: Find : FN + F = W F|| - Ff = FNET m = W= FA= F= F||= 6.12 60.0 55 31.5 45.1 FNET = ma F|| - Ff = FNET FN + F = W Ff =  FN = 6.12(3) 45.1 – Ff = 18.4 FN + 31.5 = 60 26.7=  (28.5) 18.4 N 26.7 N 28.5 N .937

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