Dinâmica

Problema 3-3

The crate of weight W has a velocity vA when it is at A. Determine its velocity after it slides down the plane to s = s'. The coefficient of kinetic friction between the crate and the plane is μk. Given: W = 20 lb vA = 12 s' = 6 ft ft s a = 3 b = 4

μ k = 0.2
Solution: a θ = atan ⎛ ⎞ ⎜ ⎟ NC = W cos ( θ ) m s v' = Find ( v' ) v' = 17.72 ft s

⎝ b⎠

F = μ k NC

Guess

v' = 1

Given
Problema 3-4

1⎛ W⎞ 2 1⎛ W⎞ 2 ⎜ ⎟ vA + W sin ( θ ) s' − F s' = ⎜ ⎟ v' 2⎝ g ⎠ 2⎝ g ⎠

A car is equipped with a bumper B designed to absorb collisions. The bumper is mounted to the car using pieces of flexible tubing T. Upon collision with a rigid barrier at A, a constant horizontal force F is developed which causes a car deceleration kg (the highest safe deceleration for a passenger without a seatbelt). If the car and passenger have a total mass M and the car is initially coasting with a speed v, determine the magnitude of F needed to stop the car and the deformation x of the bumper tubing. Units Used: Mm = 10 kg kN = 10 N Given: M = 1.5 10 kg v = 1.5 Solution: The average force needed to decelerate the car is F avg = M k g F avg = 44.1 kN 1 x = M 2 m s k = 3 The deformation is T1 + U12 = T2 1 2 M v − F avg x = 0 2 x = 38.2 mm
3 3 3

⎛ v2 ⎞ ⎜ ⎟ ⎝ Favg ⎠

Problema 3-7

The crate of mass M is subjected to forces F 1 and F2, as shown. If it is originally at rest, determine the distance it slides in order to attain a speed v. The coefficient of kinetic friction between the crate and the surface is μk. Units Used: kN = 10 N Given: M = 100 kg F 1 = 800 N F 2 = 1.5 kN v = 6 m s
3

μ k = 0.2 g = 9.81 m s
2

θ 1 = 30 deg θ 2 = 20 deg
Solution:

NC − F 1 sin ( θ 1 ) − M g + F2 sin ( θ 2 ) = 0 NC = F1 sin ( θ 1 ) + M g − F 2 sin ( θ 2 ) T1 + U12 = T2 F 1 cos ( θ 1 ) s − μ k Nc s + F2 cos ( θ 2 ) s = Mv
2

NC = 867.97 N

1 2 Mv 2

s = 0.933 m 2( F 1 cos ( θ 1 ) − μ k NC + F 2 cos ( θ 2 ) ) Problema 3-10 The block of mass M is subjected to a