Magnitude of force (elevator question)

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The discussion revolves around calculating the force exerted by the elevator floor on a person standing inside, considering the person's weight and the elevator's acceleration. The individual initially struggles with the concepts of force, acceleration, and velocity, questioning how to incorporate them into their calculations. They present an equation that combines weight, acceleration, and velocity, which they believe leads to the correct answer. However, another participant points out that the equation improperly mixes units of force and momentum, indicating a misunderstanding in the application of physics principles. The conversation highlights the importance of correctly applying Newton's laws and distinguishing between different physical quantities in problem-solving.
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Homework Statement


Pat Summit (60 kg) is standing in an elevator which has a
velocity of 10.0 m/s downward and an acceleration of 5.00
m/s^2 upward. What is the magnitude of the force exerted by
the floor of the elevator on her? Is the elevator speeding up
or slowing down?

Homework Equations


F=ma..

The Attempt at a Solution


I really am not sure where to start here..
I drew a free body diagram but, does acceleration count as a force?
How does velocity come into play?
 
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On the free body diagram, there is normal reaction and weight. So what is the resultant force 'ma' equal to?
 
I got the answer :)
using V*m + W - F = m*a_y... so..
F = (60*9.81) - (-5m/s^2)*m + (60kg * 10m/s)
thanks for the help.
 
studybug said:
I got the answer :)
using V*m + W - F = m*a_y... so..
F = (60*9.81) - (-5m/s^2)*m + (60kg * 10m/s)
thanks for the help.

While I am glad to see that you got your answer, your equation doesn't seem to 'add' up

as

60*9.81 = N

5 m/s2 * 60kg = N

60kg*10m/s = kgm/s


which means you are adding force+force+ momentum and getting force. :confused:
 
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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