Free Body Diagram Probelm Help

[pointintime]

Homework Statement

26. A box weighing 85 N rests on a table. A rope tied to the box runs vertically upward over a pully and a weight is hung form the other end. Detrmine the force that the table exerts on the box if the weight hanging on the other side of the pully weighs (a) 30 N, (b) 60 N, and (c) 90 N.

Homework Equations

net force = ma

The Attempt at a Solution

Ok well we know that the Fn of the first block is 85 N and the Fn of the other block changes so for (a) I'm not exactly sure how to do this problem...

Do I take the Fn of the first block and the Fn of the second block and then find the Fn exerted by the table

or do I take the Fn of the first block and the Fg of the second block...

not exactly sure how to do this thanks...

 

[pointintime]

do I simply find it this way

mg = -(Fn1 + Fn2)
then force exerted by table = -(mg)
?

 

[LBloom]

Remember, F_n (the normal force) is a contact force and only the first block is in contact with the table. Second F_n is not constant! It's simply the tables response to the gravitational force exerted on it by the table. A contact force does what it has to support an object, nothing more.

 

[pointintime]

wait so I'm not sure what to do...

 

[pointintime]

so it's the same for every problem... just 85 ... ? the other block doesn't matter... ? I thought the Fn of the hanging block was transferred through the rope or something...

 

[LBloom]

the force of the other block is transferred to the other block and it does matter.
However, I'm not sure wat Fn means? Is that the normal force, because there is no normal force with the second block because its not in contact with the table. Normal force, is the force the a surface exerts on the object it supports
Heres what you should do, draw a free-body diagram remember that the Fg (gravitational force of the second block) is transferred to the second block through the wire.

 

[pointintime]

Fn is the normal force I'm just nor sure how to do this problem... How do I do it isn't the the force of tension equal and opposite the force of Fg on the second block Fg is the force of gravity and Fn is the normal force...

How do I solve this problem?

 

[pointintime]

do I just apply the net force in the y direction? plus the force of tension in the wire because it's transferred thorugh the rope and also goes downwards?

net Fy = Fg1 + Fg2 + Ft
were Fg1 is the Fg of the block on the table and
Fg2 is the other block and Ft is opposite and equal Fg2 ?

 

[pointintime]

- 80 N - 30 N + 30 N
for (a) is the net force in the y direction
= - 50 N
so is the answer to problem (a) 50 N?

 

[LBloom]

your getting closer, so no need to panic.
You're right, for the second block, Ft is equal to Fg of the second block and you should just focus on the net force in the y direction.
However you should rethink your equation for net Fy. Remember, you want to focus on the Net Fy on block 1 only. Therefore the only forces there should be ones that act on that block.

 

[pointintime]

so then
- 80 N - 30 N + Ft = 0

Ft = 120 N

 

[pointintime]

so then net force in y direciton for block 1 is 0 because it's not moving... ?
net force y block 1 = - 85 N... but in the problem it's not moving is it?

 

[pointintime]

net force y block 1 = Fn + Fg
were Fg is negative

net force y block 2 = Ft + Fg
were Fg is negative
and Ft is equal and opposite to Fg

 

[LBloom]

First off, your solving for the normal force, not the tension force because they give you Ft! you said so yourself, the tension force is equal to the force of the second block and the force of tension in the rope is uniform.
There are only three forces on block 1. The tensional force, which we've already said is equal to block 2. Theres the gravitational force which is given. Finally, there is the normal force which is a contact force. It does what it needs to do to support the block and nothing more
Heres what you should do, draw a freebody diagram with two forces. One the tensional force. The other, the gravitational force. Once you've computed what the net force is from these two forces alone, you should be able to say what the normal force is

edit: your equation for net force in the y direction for block 2 is correct but unneeded. Your equation for net force in the y direction for block 1 forgets one important force.

 

[pointintime]

http://img7.imageshack.us/img7/8431/lolmlv.jpg

hmm
net force block 1 y direction = Ft + Fn + Fg = 0
were Fg is negative...

 

[pointintime]

gee thanks!
i think i got it
block 1 doens't exert a force onto the rope?

 

[LBloom]

glad to see you understand it, but be careful for the last part where block 2 weighs more than block 1

 

[pointintime]

how do i do that :O
block one would go flying up so..

 

[LBloom]

block 1 doens't exert a force onto the rope?

Nope, because its supported by the table.

 

[pointintime]

net force would still equal
Fn + Ft - Fg
but the second block would be supported by the table correct instead of the first one?

 

[LBloom]

block one would go flying up so..

exactly! If the block 1 is flying upwards and not on the table, what force does the table apply on the block?

 

[pointintime]

none...?

 

[pointintime]

but how do i prove that block 1 would go flying up..?

 

[LBloom]

the net force on block 2 is the Ft-Fg, and when the block is in contact with the table, its normal force makes up the difference. Hence: Ft+Fn-Fg=0 or Fn=Fg-Ft
However, if Ft is larger than Fg, than the required normal force to keep the block still would have to be negative, which is simply impossible because the table doesn't have any gripping power. Once Fg=Ft. there is no normal force because it is no longer needed and no extra force is being pushed on the table.

 

[LBloom]

none...?

I don't know how your teacher wants you to prove it, but yea, the normal force would be zero