The Force a man exerts on an object when in a elevator

[Amy06]

A courier is delivering a 5kg package to an office high in a tall building.
a) What upwards force does the courier apply to the package when carrying it horizontally at a constant velocity of 2m/s into the building?
B) The Courier uses the elevator to reach the office. While the elevator (Containing the courier holding the package) is accelerating at 0.11m/s/s, what upwards force does the courier apply to the package?
C) when the elevator is traveling upwards at a constant speed of 6m/s what upwards force deos the courier apply to the package?
D) in order to stop at the correct floor the elevator accelerates downwards at a rate of 0.2m/s. What is the upward force that the courier applies to the package during deacceleration?



Force= mass x acceleration
Weight = mass x gravity constant




A)

W= mg
= 5 x 9.81
= 49N upwards

(F=ma - f=5x2 f= 10N forwards)

B)

F= ma
= 5 x 0.11
= 0.55N upwards

However I am struggling with questions B,C and D as I don't know whether the 0.55N needs to be added to the 49N upwards already or whether it is just plain 0.55N. I think the courier applies the same 49N upwards and the elevator forces him up at 0.55N but I'm guessing this is wrong. Anyone know the answers or could help out with the remaining questions? Ta :)

 

[Dadface]

Hello Amy,
It will probably help you to see the reasoning behind the question if you sketched free body force diagrams:
B.There are two forces on the package
1.The weight of 49N which acts downwards and remains constant
2.The upward force(F) that the courier exerts.
Since the lift(elevator) is accelerating upwards F must be bigger than 49N and we can write:
F-49=Ma=5 times 0.11.So you do add the 0.55N to the weight but hopefully you can now see more clearly why.

 

[Lostinthought]

in question C he has to apply only 49N force as the elevator is not accelerating

 

[HallsofIvy]

In A there us NO horizontal force. That "2 m/s" is velocity not acceleration. Since his velocity is constant, his acceleration is 0.

In B, the elevator is pushing him upward at 0.11 m/s/s and the courier must then apply f= ma to the package in addition to its weight.

 

[Amy06]

Yeah that does make it clearer, Thank you!
In D) I just use, F=ma= 5 x -0.2= 1N downwards
and then add the 49N up and -1N and get 48N upwards force when the elevator is deaccelerating. Correct?

 

[Dadface]

Right answer but try to get into the habit of sketching free body force diagrams and writing out the second law showing the resultant force:

49-F=Ma

 

[Amy06]

Ok yeah, I see the benefit now :) Thank you so much!