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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. Finally, it reminds the reader of the equations to calculate weight as the product of density, volume, and gravitational acceleration.

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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