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توازن جسم صلب خاضع لقوتين

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Karim Annouh
Karim Annouh

This document discusses the equilibrium of a solid body subjected to two forces. It explains that the tension T in a spring is directly proportional to the extension Δl of the spring, following the equation T = KΔl, where K is the spring constant. It then provides an example where a body is suspended by two springs, with tensions of 2N and 1.4N respectively, resulting in a net force of 0.6N acting on the body. The force is equal to the weight of the liquid displaced.

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Equilibre d’un corps solide soumis à deux forces  .1 .2 .I    300 4.0 250 3.5 200 3.0 150 2.5 100 2.0 50 1.5 0 1 P=T 3.0 3.0 2.5 2.5 2.0 2.0 1.5 1.5 m (g) l (cm) (S) 1.0 1.0 0.5 0.5 0 0 T (N) ∆l (cm) .1 .2
.3 T = K.∆l K .II ∆l T .1 ∆l T = K.∆l N.m-1 (S.I) K T = K.∆l  (S) T2 = 1.4N .1 T1 = 2N Fa = T1 – T2 = 0.6N (S) .2
V = 58cm3 (S) Pliquide = ρ.V.g Fa ≈ Pliquide V Pliquide = 0.58N .III .1 .2     F = ρ.V.g Kg.m-3 m3 N.Kg-1 ρ V g

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توازن جسم صلب خاضع لقوتين

  • 1. Equilibre d’un corps solide soumis à deux forces  .1 .2 .I    300 4.0 250 3.5 200 3.0 150 2.5 100 2.0 50 1.5 0 1 P=T 3.0 3.0 2.5 2.5 2.0 2.0 1.5 1.5 m (g) l (cm) (S) 1.0 1.0 0.5 0.5 0 0 T (N) ∆l (cm) .1 .2
  • 2. .3 T = K.∆l K .II ∆l T .1 ∆l T = K.∆l N.m-1 (S.I) K T = K.∆l  (S) T2 = 1.4N .1 T1 = 2N Fa = T1 – T2 = 0.6N (S) .2
  • 3. V = 58cm3 (S) Pliquide = ρ.V.g Fa ≈ Pliquide V Pliquide = 0.58N .III .1 .2     F = ρ.V.g Kg.m-3 m3 N.Kg-1 ρ V g