Taller No 2 Dinamica Grupo Ad, Listo-1.docx

WORKGROUP No. 2 KYNETICS OF PARTICLE CIVIL ENGINEERING ENG. JORGE LUIS PACHECO YEPES, M.Sc.. 1. The 800-kg car at B is connected to the 350-kg car at A by a spring coupling. Determine the stretch in the spring if (a) the wheels of both cars are free to roll and (b) the brakes are applied to all four wheels of car B, causing the wheels to skid. Take(𝝁𝒌 )𝑩 = 𝟎. 𝟒. Neglect the mass of the wheels.

Solución a) ∑ Fx = ma ∑ (FxA + FxB) = (mA + mB) a 4 5

(800 + 350)(9,81) = (800 + 350) a

a= 7,848 m/s2 Debido a que están conectados tienen la misma aceleración Para el carro A ∑ Fx = ma 4 5

(350)(9,81) - Fs = (350)( 7,848) 4

Fs = 5 (350)(9,81) - (350)( 7,848) = 0 Fs = k × x Siendo x “stretch” X=

Fs 𝑘

=0

X=0 El resorte no se estira b)

4 5

3

(800 + 350)(9,81) – (0,4)(800)(9,81) 5 = (800 + 350) a

a= 6,21 m/s2 4 5

(350)(9,81) - Fs = (350) (6,21) Fs = 573,25 N

X=

573,25 600

= Fs = 0,955 m

2. The ball has a mass m and is attached to the cord of length L. The cord is tied at the top to a swivel and the ball is given a velocity V0. Show that the angle θ which the cord makes with the vertical as the ball travels around the circular path must satisfy the equation 𝐭𝐚𝐧(𝜽) 𝐬𝐢𝐧(𝜽) =

𝒗𝟐𝟎 𝒈𝒍

. Neglect air resistance and the size of the ball.

Solución ∑ Fy = 0 T (Cos θ) - mg = 0 𝑚𝑔 T= 𝐶𝑜𝑠Ꝋ ∑ Fn = man T (Sen θ)= m

𝑉02 𝑅 𝑉 2

𝑚𝑔

0 (Sen θ) = m 𝐿 (𝑆𝑒𝑛Ꝋ) 𝐶𝑜𝑠Ꝋ

g(tan θ)(Sen θ)= (tan θ)(Sen θ) =

𝑉 02 𝐿 𝑉 02 𝑔𝐿

3. The 0.75-lb smooth can is guided along the circular path using the arm guide. If the arm has an angular velocity 𝜽̇ = 𝟐 𝒓𝒂𝒅/𝒔 and an angular acceleration 𝜽̈ = 𝟎. 𝟒 𝒓𝒂𝒅/𝒔𝟐 at the instant 𝜽 = 𝟑𝟎º, determine the force of the guide on the can. Motion occurs in the vertical plane.

Solución r = 𝐶𝑜𝑠𝜃|Ꝋ = 30º = 0.8660 ft 𝑟̇ = −𝑆𝑒𝑛𝜃𝜃̇|Ꝋ = 30º = -1,00 ft/s 𝑟̈ = −(𝐶𝑜𝑠𝜃𝜃̇ 2 + 𝑆𝑒𝑛𝜃𝜃̈)|Ꝋ = 30º = -3,664 ft/s2 ar = 𝑟̈ - r𝜃̇ 2 = -3,664 - 0.8660(2)2 = -7,128 ft/s2 aꝊ = r𝜃̈ + 2𝑟̇ 𝜃̇ = 0.8660(4) + 2(-1)(2) = -0,5359 ft/s2 ∑ Fr = mar 0,75

-N (Cos 30º) – 0,75 (Cos 60º) = 32,2 (-7,128) N = -0,2413 lb ∑ FƟ = maƟ 0,75

F + 0,2413 (Sen 30º) – 0,75 (Sen 60º) = 32,2 (-0,5359) F = 0,516 lb