of 1/1
0121 Lecture Notes - An Introductory Tension Force Problem.docx page 1 of 1 Flipping Physics Lecture Notes: An Introductory Tension Force Problem m hanging = 155.0g × 1 kg 1000g = 0.155 kg; θ = 28° ; F T 1 = ?; F T 2 = ? sin θ = O H = F T 1y F T 1 F T 1y = F T 1 sin θ & cosθ = A H = F T 1x F T 1 F T 1x = F T 1 cosθ F y = F T 1y F g = ma y = m 0 () = 0 F T 1y F g = 0 F T 1y = F g F T 1 sin θ = mg F T 1 = mg sin θ = 0.155 ( ) 9.81 ( ) sin 28 ( ) = 3.238854 3.2N F x = F T 2 F T 1x = ma x = m 0 () = 0 F T 2 F T 1x = 0 F T 2 = F T 1x = F T 1 cosθ F T 2 = 3.238854 ( ) cos 28 ( ) = 2.85974 2.9N Note: The mass hanging is an object in translational equilibrium because the net force acting on it equals zero.

0121 Lecture Notes - An Introductory Tension Force Problem...0121 Lecture Notes - An Introductory Tension Force Problem.docx page 1 of 1 Flipping Physics Lecture Notes: An Introductory

  • View
    2

  • Download
    0

Embed Size (px)

Text of 0121 Lecture Notes - An Introductory Tension Force Problem...0121 Lecture Notes - An Introductory...

  • 0121 Lecture Notes - An Introductory Tension Force Problem.docx page 1 of 1

    Flipping Physics Lecture Notes: An Introductory Tension Force Problem

    m

    hanging=155.0g × 1kg

    1000g= 0.155kg;θ = 28°; F

    T1= ?; F

    T2= ?

    sinθ = OH

    =F

    T1y

    FT1

    ⇒ FT1y

    = FT1

    sinθ &

    cosθ = AH

    =F

    T1x

    FT1

    ⇒ FT1x

    = FT1

    cosθ

    F

    y∑ = FT1y − Fg = may = m 0( ) = 0⇒ FT1y − Fg = 0⇒ FT1y = Fg ⇒ FT1 sinθ = mg

    ⇒ FT1= mg

    sinθ=

    0.155( ) 9.81( )sin 28( ) = 3.238854 ≈ 3.2N

    F

    x∑ = FT2 − FT1x = max = m 0( ) = 0⇒ FT2 − FT1x = 0⇒ FT2 = FT1x = FT1 cosθ

    ⇒ F

    T2= 3.238854( )cos 28( ) = 2.85974 ≈ 2.9N

    Note: The mass hanging is an object in translational equilibrium because the net force acting on it equals zero.