Simple equilibrium problem

[tim51]

Homework Statement

[PLAIN]http://img651.imageshack.us/img651/8828/capage1.jpg [Broken]

Homework Equations

F = ma
Sum of moments about a point = 0

The Attempt at a Solution

My attempt can be seen above. The only part which I seem to have a problem with is part f. In order for there to be a deceleration there must be a nett force acting upwards.

Therefore I figure that part a force would remain the same as the mass hasn't changed.

part b force would also remain the same

and part c force would increase so as to produce the decelleration? f=ma so the nett force would need to be f = 600N, therefore it would increase by 600N?

 

[PeterO]

Homework Statement

[PLAIN]http://img651.imageshack.us/img651/8828/capage1.jpg [Broken]

Homework Equations

F = ma
Sum of moments about a point = 0

The Attempt at a Solution

My attempt can be seen above. The only part which I seem to have a problem with is part f. In order for there to be a deceleration there must be a nett force acting upwards.

Therefore I figure that part a force would remain the same as the mass hasn't changed.

part b force would also remain the same

and part c force would increase so as to produce the decelleration? f=ma so the nett force would need to be f = 600N, therefore it would increase by 600N?

No. Part A force has to not only match the weight, but in addition provide an upward acceleration. since that acceleration is equal in magnitude to the acceleration due to gravity when an object is dropped, the force will be twice the size.

ie: for a 500N person, the upward force of 500N means they don't move [as when you are standing there.
An upward force of 1000N is needed to get an upward acceleration of 10 ms^-2.

think: downward force 500N [gravity] upward force 1000N applied force. Net force 500N up.

while the person was falling, the 500N downward force was providing an acceleration of 10 ms^-2 down - like all things that fall.

The net force of 500N up will provide an acceleration of 10 ms^-2 up, as required.