lecture_33_web_2007

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Unformatted text preview: Quiet please ladies and gentlemen The lecture is about to begin n Statics/Equilibrium For an object to be static it is necessary for: the net force on an object to be zero And the net torque on the object is zero x F = ∑ τ = ∑ We can use these three conditions to find the forces at different points on an object in equilibrium y F = ∑ θ A beam of mass M = 280 kg and length l = 2.2m is attached to a wall as shown on the figure, It is supported by a wire making angle of θ = 30° relative to the horizontal. A Lamp of mass m = 25 kg is hanging at the end of the beam. Q. What is the force acting on the hinge? Equilibrium: problem solving x F = ∑ ( 29 cos Hx F T θ- = Mg T mg H F Hx F Hy F ϕ We need to find both the magnitude and the direction of F ! ( 29 ( 29 cos 3 Hx F T θ = x y θ Equilibrium: problem solving Mg T mg H F Hx F Hy F ϕ ( 29 ( 29 cos 1 Hx F T θ = ( 29 sin Hy T F Mg mg θ +-- = ( 29 ( 29 ( 29 sin 2 Hy F M m g T θ = +- τ = ∑ 2 l Mg ( 29 ( 29 1 2 3 sin Mg mg T θ + = x y y F = ∑ ( 29 sin T θ mgl + ( 29 sin Tl θ- = ( 29 sin 2 l Mg mgl Tl θ + = + A beam of mass M = 280 kg and length l = 2.2m is attached to a wall as shown on the figure, It is supported by a wire making angle of θ = 30° relative to the horizontal. A Lamp of mass m = 25 kg is hanging at the end of the beam. Q. What is the force acting on the hinge? θ Equilibrium: problem solving Mg T mg H F Hx F Hy F ϕ ( 29 ( 29 cos 1 Hx F T θ = ( 29 ( 29 ( 29 sin 2 Hy F M m g T θ = +- ( 29 ( 29 1 2 3 sin Mg mg T θ + = ( 29 ( 29 ( 29 1 280 9.8 25 9.8 2 sin 30 T + = 3234 N = ( 29 3234cos 30 Hx F = 2800 N = ( 29 ( 29 ( 29 280 25 9.8 3234sin 30 Hy F = +- 1372 N = + A beam of mass M = 280 kg and length l = 2.2m is attached to a wall as shown on the figure, It is supported by a wire making angle of θ = 30° relative to the horizontal. A Lamp of mass m = 25 kg is hanging at the end of the beam....
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