Find the forces on a table that is lifted in a tilted position

[Helly123]

Homework Statement

Homework Equations

The Attempt at a Solution

what's the Fa and Fb?

my logic is
Fa = Wa - N = mg - $\frac{N}{\cos \theta}$

so is Fb...
so the ratio is 1. But I know it's wrong. What will make the difference between Fa and Fb?
can anyone give a clue please?

 

[kuruman]

Clearly F_A + F_B = W. To find what fraction of W is each force you need another equation. Consider the sum of torques acting on the box.

 

[Helly123]

Clearly F_A + F_B = W. To find what fraction of W is each force you need another equation. Consider the sum of torques acting on the box.

My logic is there are 3 torques

Torque at A
Normal force*L (clockwise)

Torque at B
Normal force*L (counter clockwise)

Torque for W
mg cos thetha*L

F_A + F_B = W.

What should i do next?
Is the w do torque?
Is it necessary to find the w torque? Is the length for W torque is L?

 

[haruspex]

My logic is there are 3 torques

Generally, a torque is only meaningful about a stated axis. To combine torques in an equation they must all be using the same axis.
What point are you taking as axis? Do all the forces have a torque about that point? (Often it helps to choose an axis that is in the line of action of one of the unknown forces, so that force does not appear in the equation.)

Normal force*L (clockwise)

Look carefully at the diagram. In what direction are F_A and F_B?

 

[Helly123]

To combine torques in an equation they must all be using the same axis.

I use the table surface as the axis. So the Normal force already perpendicular to it. And W cos theta perpendicular to it.
Is it right?

Look carefully at the diagram. In what direction are F_A and F_B?

Pointing up. Perpendicular to ground. And parallel to W.I am confused to choose which point as point load for the W.
Is it necessary to find point load? Can you give me a clue?

 

[haruspex]

I use the table surface as the axis.

The axis will be a single line, viewed end-on in the diagram, so appearing as a point.
Obvious choices are A, O, B or G.

Perpendicular to ground.

So not normal to the table top, yet you mention "normal force".
Do you understand how to calculate the torque of a force about an axis?

which point as point load for the W.

The mass centre is shown as point G.

 

[Helly123]

The axis will be a single line, viewed end-on in the diagram, so appearing as a point.
Obvious choices are A, O, B or G.

So not normal to the table top, yet you mention "normal force".

Wait. I meant the Normal force perpendicular to the table. So is normal. Lol

 

[haruspex]

Wait. I meant the Normal force perpendicular to the table. So is normal. Lol

Where in the diagram is there a force normal to the tabletop?

 

[Helly123]

Where in the diagram is there a force normal to the tabletop?

No.. i guess i made it up myself.. why there is no Normal Force? So only Fa Fb and W?

 

[haruspex]

So only Fa Fb and W?

Yes.
Of course, each has a component that's normal to the tabletop.
Please choose an axis.

 

[kuruman]

Please choose an axis.

More to the point, choose a reference point about which you will find expressions for the torque. What is a good choice for such a point?

 

[Helly123]

This as far as my brain works. Can you give me clues?

 

[kuruman]

If your origin is as indicated by the arrow in your drawing then
(a) The signs of the torques are inconsistent with one another. Follow whatever convention you wish, but torques that tend to rotate the box in the same direction must have the same algebraic sign. Torques do not necessarily have positive or negative signs because the forces that generate them are positive (up) or negative (down).
(b) Recheck the lever arms that are associated with the torques. One of them is incorrect.

 

[Helly123]

If your origin is as indicated by the arrow in your drawing then
(a) The signs of the torques are inconsistent with one another. Follow whatever convention you wish, but torques that tend to rotate the box in the same direction must have the same algebraic sign. Torques do not necessarily have positive or negative signs because the forces that generate them are positive (up) or negative (down).
(b) Recheck the lever arms that are associated with the torques. One of them is incorrect.

how about this? Sorry for the rotation. I cannot change it

 

[kuruman]

The torque equation is correct. However, your answer should not have F_B or mg on the right side otherwise you will not be able to select the correct choice from the given 6. You forgot the force equation. Use it to eliminate W from the torque equation, then find the ratio.

If you post another photo, please make sure that there is plenty of light and use a dark pen to increase contrast. The rotation doesn't bother me because I can turn my laptop 90^o to look at it.

 

[haruspex]

More to the point, choose a reference point about which you will find expressions for the torque. What is a good choice for such a point?

That is what I mean by choosing an axis.

 

[kuruman]

That is what I mean by choosing an axis.

Yes, I know. I wanted to clarify the point (pardon the pun) for OP who wrote in post #5, "I use the table surface as the axis."

 

[haruspex]

Yes, I know. I wanted to clarify the point (pardon the pun) for OP who wrote in post #5, "I use the table surface as the axis."

Ok, but I thought I already clarified that in post #6.

 

[kuruman]

Ok, but I thought I already clarified that in post #6.

You did ... and I missed it.

 

[Helly123]

I don't understand.. using right hand rule. Which direction W torque go?

 

[kuruman]

I don't understand.. using right hand rule. Which direction W torque go?

As usual, if the fulcrum is to the left of a vertical line passing through the point of application of W, the torque is counterclockwise; if the fulcrum is to the right of the line, the torque is clockwise.

On edit: This is incorrect. Please see my correction below in post #23.

 

[Helly123]

As usual, if the fulcrum is to the left of a vertical line passing through the point of application of W, the torque is counterclockwise; if the fulcrum is to the right of the line, the torque is clockwise.

W is more like at the middle btw. And W force is going down

 

[kuruman]

Look at your drawing in post #12. Imagine a vertical line through the fulcrum. Is W to the right, to the left or right on the vertical line?

On Edit: I just realized that I mistakenly interchanged my directions for down forces. The correct rule is Fulcrum to the left, force down, torque clockwise. Fulcrum to the right, force down, torque counterclockwise. Sorry about the confusion.

 

[Helly123]

Look at your drawing in post #12. Imagine a vertical line through the fulcrum. Is W to the right, to the left or right on the vertical line?

On Edit: I just realized that I mistakenly interchanged my directions for down forces. The correct rule is Fulcrum to the left, force down, torque clockwise. Fulcrum to the right, force down, torque counterclockwise. Sorry about the confusion.

It is ok. Kind more make sense now. Thank you

 

[Helly123]

Um.. How about choosing A as the axis?

 

[kuruman]

Um.. How about choosing A as the axis?

You can choose any point as your axis, A, B, O, G or any other point you wish. If the system is in static equilibrium the sum of torques acting on it about all axes will be zero. It will not suddenly acquire a net torque because of your choice of axes. Therefore, it makes sense to pick an axis that makes the torque calculation as easy as possible to you. If you pick O, you have to calculate three torques, but the lever arms are easier to find; if you pick A, B or G, you only have two torques to calculate, but finding the lever arms may be a bit more involved. Any choice will give the same answer.

 

[Helly123]

You can choose any point as your axis, A, B, O, G or any other point you wish. If the system is in static equilibrium the sum of torques acting on it about all axes will be zero. It will not suddenly acquire a net torque because of your choice of axes. Therefore, it makes sense to pick an axis that makes the torque calculation as easy as possible to you. If you pick O, you have to calculate three torques, but the lever arms are easier to find; if you pick A, B or G, you only have two torques to calculate, but finding the lever arms may be a bit more involved. Any choice will give the same answer.

yes. so, the torque for $F_A$ = 0
left is torque $F_B$ and torque W
lever arms for torque $F_B$ = 2Lcos$\theta$
lever arms for W = Lsin$\theta$
both torques have the same directions.

sigma torque = 0
-torque $F_B$ = torque W
-$F_B$* 2Lcos$\theta$ = ( $F_B + F_A$)* Lsin$\theta$
( $F_B ) / (F_A$) = ( -2Lcos$\theta$ - Lsin$\theta$ ) / Lsin$\theta$
( $F_B ) / (F_A$) = ( -2cos$\theta$ - sin$\theta$ ) / sin$\theta$
( $F_B ) / (F_A$) = ( -2$\theta - tan\theta ) / tan\theta$

which is wrong sir?

 

[kuruman]

The distance from A to G where gravity acts is $L\sqrt{2}$, not $L$. Also the appropriate trig function should involve $45^o-\theta$. Make a drawing and you will see.

 

[Helly123]

Can you elaborate it? Why there is 45-$\theta$ ? (Sorry for the pic)

 

[kuruman]

Look at the picture below. (I just noticed that my drawing has points A and B interchanged, sorry). You want the angle $\varphi$ that is formed with distance AG (red line) and the perpendicular to it (dotted red line). Then the torque due to the weight is $\tau_W=W (AG) \cos\varphi$. To find $\varphi$ note that when the table is rotated back to horizontal, the acute angle between the weight and AG becomes 45^o. Apply similar reasoning to your picture which has points A and B interchanged. Hint: The answer will be slightly different from this picture.