Calculating Tension in a Suspended Loudspeaker

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To calculate the tension in the cables supporting a 15.0 kg loudspeaker suspended 1.20 meters below the ceiling, a free-body diagram is essential. The speaker is assumed to be in equilibrium, meaning the net force acting on it is zero. By analyzing the forces and applying geometry and trigonometry, one can determine the tension in each cable. Understanding the units of measurement is crucial for accurate calculations. This approach provides a solid foundation for solving the problem effectively.
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A 15.0 loudspeaker is suspended 1.20 below the ceiling by two 3.50 long cables that angle outward at equal angles. what is the tension in each cable?
im really weak at this topic..
cud any1 give me a lift off here? :frown:
 
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I presume that's a 15.0 kg speaker, and the distanced you give are in meters? The units are important.

Anyway - do a free-body diagram of the speaker. That'll give you the vectors you need to add. We'll assume that the speaker is not accelerating (since otherwise the cables have probably broken), so what can you say about the net force on it? That'll give you the vector sum of the forces in your diagram, and from that point on it's just geometry and trigonometry.

Does that help?
 
it sure did..ty
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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