Moment of force, stabilising a board carried over shoulder

[Rbraind]

Homework Statement

A 3m board is carried and has a weight of 32N. 1.8m sticks out behind the carriers shoulder,1.2m in front and his hand is 0.4m from the shoulder.How much downward force needs to be applied at the hand to stabilise the board. Assume the weight is evenly distributed along the board. I'm confused.

Homework Equations

T=Fd

The Attempt at a Solution

how do i measure the force over the whole board, will it be 32N at 1.8m and 1.2m and 0.4m?

 

[tiny-tim]

Hi Rbraind!

There are three forces on the board: the weight, and the two reaction forces.

Do vertical components, then do moments (about some suitable point) …

what do you get? ​

 

[Rbraind]

thankyou tiny-tim! - not sure if I'm completely on the right track but am feeling warmer.
Three forces on the board - the 32N and then his hand and shoulder will have reaction forces yes? So moments would be moment of inertia I am thinking I=mr2
If I do 1.5 (the some point?...pivot point!) and x by 0.4m I get 0.24, then do I times this by gravity 9.81 ?? We only covered this really briefly in our lecture so I'm trying to make sense of my textbook...hope I'm getting closer. Can't wait till I can be the one giving good advice on the forums...one day :)

 

[tiny-tim]

If I do 1.5 (the some point?...pivot point!) and x by 0.4m I get 0.24, then do I times this by gravity 9.81 ??

(moment of inertia is irrelevant)

Call the reaction forces A and B.

Write out the vertical components equation, and then the moments equation about the shoulder as the pivot point.

 

[Rbraind]

Nope I'm lost. But thanks for your help.I feel like its in there but the more i try to make sense of it the more my brain feels like exploding. 32N pushes down on his shoulder and 32N back up - pivot point must be 0.3m from his shoulder and 0.7m from his hand ? I don't get vertical components and moment equation ? I'm studying by distance and our lecture really skimmed through this bit, i should take it up with the lecturer but think I've left it too late. I love trying to work it out but I just can't make sense of the method...arggghhhh

 

[gneill]

Perhaps a diagram would help.

 

[tiny-tim]

Hi Rbraind!

32N pushes down on his shoulder and 32N back up

That's right … so the vertical components equation is A + B = 32

- pivot point must be 0.3m from his shoulder and 0.7m from his hand ? I don't get vertical components and moment equation ?

Yes and no

You can choose the pivot point to be anywhere.

You've chosen it to be where the weight W is.

So you'll have a moments equation involving A and B but not W.

That's fine … that'll be your second equation in A and B, and you can solve the two equations simultaneously.

Perhaps an easier way would be to use the shoulder as the pivot point … then your moments equation involves A and W but not B, which is quicker to solve.

Do it both ways, just to see …

what do you get? ​