Calculating Force Needed to Pull Chain Onto Table

[goonking]

Homework Statement

Homework Equations

W = F d
F = ma

The Attempt at a Solution

so in order to get the whole chain on the table, we need to pull the chain 0.65 meters onto the table.

since 0.65 meters is hanging off the table, the gravity is acting on it, therefore F=ma where m is half the chain (8kg /2 = 4kg) and a = gravity(g), which is 39.2 N,since we work against this force we need to find the force required to pull 8 Kg a distance of 0.65m and then subtract 39.2 N.

is this the correct approach?

 

[CWatters]

As you pull it up the force required get less. I would use a different approach.

Hint: Work = change in energy

 

[goonking]

As you pull it up the force required get less. I would use a different approach.

Hint: Work = change in energy

W = KEf - KEi

KEf = 0 since v =0

KEi = mgh = 8 kg x 9.8m/s^2 x 0.65 = 51

W = 51 Js

is this correct?

 

[CWatters]

Close.
How much mass changes height?
Where is the centre of that mass and how much height does it gain?

 

[goonking]

Close.
How much mass changes height?
Where is the centre of that mass and how much height does it gain?

I didn't learn about center of masses yet , is that required to do this problem? :(

 

[CWatters]

There is another way but I suspect its a lot harder.

Read up on the centre of mass or centre of gravity.

Consider a uniform ruler that's 12" long. To find the centre of mass you could balance it on a knife edge to make a seesaw (tetter-totter in the USA). If you adjust the position of the ruler until it balances the centre of mass will be found to be around the 6" position.

The centre of mass of the part hanging over the edge would be at 0.65/2 = 0.325m below the top of the table.

 

[goonking]

There is another way but I suspect its a lot harder.

Read up on the centre of mass or centre of gravity.

Consider a uniform ruler that's 12" long. To find the centre of mass you could balance it on a knife edge to make a seesaw (tetter-totter in the USA). If you adjust the position of the ruler until it balances the centre of mass will be found to be around the 6" position.

The centre of mass of the part hanging over the edge would be at 0.65/2 = 0.325m below the top of the table.

then wouldn't we need to find the center of mass for .325m and so forth?

 

[CWatters]

then wouldn't we need to find the center of mass for .325m and so forth?

No.

The part of the chain that hangs over the edge can be replaced/modelled by a single 4kg mass on a rope that hangs 0.352m over the edge.

 

[goonking]

No.

The part of the chain that hangs over the edge can be replaced/modelled by a single 4kg mass on a rope that hangs 0.352m over the edge.

so the work required to pull the lingering half of the chain up is 4kg x 9.8 x .352m?

and the work required to pull the other half which is already on the table is : W = F d

we just need to add (F d) + (4kg x 9.8 x 0.352m) right?

 

[CWatters]

so the work required to pull the lingering half of the chain up is 4kg x 9.8 x .352m?

Correct.

and the work required to pull the other half which is already on the table is : W = F d

No. It actually takes no energy/work to move that bit along the table. The problem states you can ignore friction so the "F" in W = Fd is zero.

 

[goonking]

Correct.
No. It actually takes no energy/work to move that bit along the table. The problem states you can ignore friction so the "F" in W = Fd is zero.

ok, i got the answer to be 12.7 j

 

[CWatters]

I agree.