Tension in Ropes Equal angles DO Now Mass3

Tension in Ropes: Equal angles

DO Now: Mass=3. 0 kg Fapplied = 40. 0 N Θ=75°

RKS

1) kg, m, s 2)Equations: a) Linear: Y=3 x + 2 b) Quadratic: y= 2 x 2 + 5 x + 2 c) Exponential: y = 4 x d) Inverse: y = 2/x 3) (a) +4 m/s (d) -3 m/s 2

4) Distance, Displacement: a) 6 m, 3 m east b) 15 m, 13 m approximately northeast c) 10 m, 6 m south 5) Speed, Velocity a) 3 m/s, 1. 5 m/s east b) 7. 5 m/s, 6. 5 m/s approximately east c) 5 m/s, 3 m/s south

6. Units of velocity? m/s 7. Units of acceleration? m/s 2 8. Units of force? N or mkg/s 2 9. Units of distance? m 10. Formula for average acceleration? Change in velocity/time 11. Formula for average velocity? Change in displacement/time 12. Formula for average speed? Change in distance/time 13. What is the constant acceleration of gravity on earth? -10 m/s 2 14. One type of field force? gravity 15. Three trig function ratios? Sin = O/H Cos = A/H Tan = O/A

Q U I D S FOR M Tension in each side of the wire = FORCE N or kg m/s 2 m= 2. 0 kgm, Fgrav = -10 m/s 2 2 Fnet = 0 = FT 1 + FT 2 - Fgrav Sin=opp/hyp, cos=adj/hyp F = ma 1) Fgrav = (2. 0 kg) (-10 m/s 2) = -20 kg m/s 2 2)Ty=opposite from theta T = hypotenuse Sin θ= Ty/T 3) Then I can solve for ty= T Sin θ 4) I can substitute everything I know now: Fnet = - Fgrav + FT 1 + FT 2 0 =-20 kg m/s 2 +Tsin(10) 20 kg m/s 2 = 2 T(0. 17) 10 kg m/s 2 = 0. 17 T T= 58. 8 kg m/s 2 E T = 60 N

Q U I D S FOR M E