Elevator Tension Calculation: Finding Force with Fnet and Newton's Laws

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To calculate the tension in the cable of an elevator with a mass of 650 kg accelerating upward at 3.00 m/s², the net force (Fnet) is determined using Fnet = ma, resulting in 1950 N. The weight of the elevator (W) is calculated as W = mg, where g is 9.8 m/s², giving W = 6370 N. The tension (T) in the cable must counteract both the weight and provide the upward acceleration, leading to the equation T - W = Fnet. Thus, the tension can be found using T = W + Fnet. Understanding these relationships is crucial for solving tension problems in physics.
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Homework Statement



An elevator with a mass of 650 Kg supported by steel cable. What is the tension in the calbe when the elevator is accelerated upward at the rate of 3.00 m/s^2


Homework Equations



well he gives the equations that we can use as: Fnet=ma, w=mg and g=9.8m/s^2
I don't know what W means. Also he never taught us how to find tension.


The Attempt at a Solution



Well I used Fnet: 650KG * 3.00m/s^2= ANS.

If this is right can someone explain to me how.


x

Please Help.
 
Last edited:
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t00kool said:

The Attempt at a Solution



Well I used Fnet: 650KG * 3.00m/s^2= ANS.

If this is right can someone explain to me how.

well the resultant acceleration is 3 ms^{-2}
so the net force,F_{net}=650*3=1950N

The weight of the elevator is W.
The tension in the cables holding the elevator is T.

If the weight is acting down and tension is acting up, and the resultant of these 2 is in an upward direction. What would be an equation relating the the resultant (net) force, the tension and weight?
 
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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