Horizontal force by block on a beam

[hello478]

Homework Statement

The block is now moved closer to end A of the beam. Assume that the beam remains
horizontal.
State and explain whether this change will increase, decrease or have no effect on the horizontal
component of the force exerted on the beam by the hinge

Relevant Equations

vectors

 

[BvU]

9.2 ? Ah, $W$

$\tau = 0$ and $\tau = 12*0.35+W*0.25-T_y * 0.5 \ne 0$ at the same time ?

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[hello478]

9.2 ?

9.2 is weight of beam as mentioned on the diagram...

 

[BvU]

I appear to be looking at the tail end of something.
Was the 17 N given, or is it the rounded off result of an equilibrium calculation ?

PS can you type instead of snapping cue cards ?
A complete problem statement would be nice too
And: am I missing your question ?

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[hello478]

I appear to be looking at the tail end of something.
Was the 17 N given, or is it the rounded off result of an equilibrium calculation ?

PS can you type instead of snapping cue cards ?
A complete problem statement would be nice too
And: am I missing your question ?

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17 was given in the question
here is the question...

 

[hello478]

PS can you type instead of snapping cue cards ?
A complete problem statement would be nice too
And: am I missing your question ?

$\$

yeah ill try to type next time
and my question is what would all the forces on the hinge be? and how to calculate them?
and why would the horizontal force at hinge decrease by moving block a closer to end A of beam

 

[BvU]

So at this late stage I have to discover that $W$ was not given, but calculated ?

You probably answered a with $\sum \vec F = \vec 0$ and $\sum \vec \tau = \vec 0$ ...

And b with $T\sin\Bigl ({50\over 180}\pi\Bigr ) = 13.023 \approx 13$ N

In c you calculate 9.25 N from $\sum \vec \tau = \vec 0$

d uses $\sum F_y = 0$ to get 8.22 N (upward) at the hinge

and e doesn't require you to do any calculating. But $\sum F_x = 0$ tells you the horizontal force the hinge exerts on the beam.
Moving the block does change all three balances and qualitatively you should be able to say that $T$ decreases, and therefore the horizontal force the hinge exerts on the beam decreases as well.

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[hello478]

So at this late stage I have to discover that $W$ was not given, but calculated ?

You probably answered a with $\sum \vec F = \vec 0$ and $\sum \vec \tau = \vec 0$ ...

And b with $T\sin\Bigl ({50\over 180}\pi\Bigr ) = 13.023 \approx 13$ N

In c you calculate 9.25 N from $\sum \vec \tau = \vec 0$

d uses $\sum F_y = 0$ to get 8.22 N (upward) at the hinge

and e doesn't require you to do any calculating. But $\sum F_x = 0$ tells you the horizontal force the hinge exerts on the beam.
Moving the block does change all three balances and qualitatively you should be able to say that $T$ decreases, and therefore the horizontal force the hinge exerts on the beam decreases as well.

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yeah you got everything right
thanks to your amazing telepathic powers you guessed my answers correct too
but i dont understand part e quite well
can you please expand more on that? like in simple words...

 

[BvU]

but i dont understand part e quite well
can you please expand more on that? like in simple words

Move the block to the right $\Rightarrow$ need bigger $T_y$ to compensate for the torque. Angle stays the same $\Rightarrow$ $T_x$ bigger too. Only other horizontal force is from the hinge.

Bigger $T_y$ from the string $\Rightarrow$ less upward force needed from the hinge for the $\sum F_y = 0$ balance.


Conversely:
Move the block to the left (towards the hinge) $\Rightarrow$ need less $T_y$ to compensate for the torque. Angle stays the same $\Rightarrow$ $T_x$ smaller too. Only other horizontal force is from the hinge $\Rightarrow$ less horizontal force (to the left) required from the hinge.

Extreme case: block at point A.

Test to see if you understand it now: what is the horizontal component of the force that the hinge exerts on the beam when the block is at point A ? And the vertical component ?

Optional: same question for block at point B.


Checking:
more

and simple

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[hello478]

Checking:
more View attachment 342408and simple View attachment 342409

i understood it a bit
thank you!

 

[BvU]

Test to see if you understand it now: what is the horizontal component of the force that the hinge exerts on the beam when the block is at point A ? And the vertical component ?

 

[hello478]

at point A
the horizontal force would be =
and the vertical force would be =
i cant seem to solve it

 

[BvU]

Ok, so we have some more work to do. First thing is to make a sketch

Now, that wasn't too difficult, right ?

Next: write down the three balances.
List and check the variables
Check which are knowns and which are unknown.
If necessary, add more equations ...
You want to end up with just as many equations as you have unknowns

Then: solve

Post your work so we can comment and provide guidance

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[hello478]

Then: solve

Post your work so we can comment and provide guidance

hy = 12 + x
w= 9.2
ty = 17 sin 50 = 13
so x would be 3.8 upwards
so hy = 15.8

and hx 17 cos 50 = 10.9 in opposite direction to tx
but wouldnt the horizontal component remain same always then?

 

[BvU]

write down the three balances

$\sum \tau = 0 \Rightarrow T_y * 0.5 = W*0.25$
$\sum F_x = 0 \Rightarrow T_x + H_x = 0$
$\sum F_y = 0 \Rightarrow T_y + H_y + W + 12 N = 0$

List and check the variables

Check which are knowns and which are unknown

I filled in the 0.5 and 0.25 m distances. W is also known.
Unknowns are $H_x, H_y, T_x, T_y$ (the first two are pretty obvious, but $T$ also changes ! -- which I wanted you to discover by following the steps I wrote down)

So:

wouldnt the horizontal component remain same always then?

No! The 17 N was when the block was at 0.35 m.

If necessary, add more equations ...
You want to end up with just as many equations as you have unknowns

$T_x = T\cos\bigl ( {50\over 180}\pi \bigr )$ and $T_y = T\sin\bigl ( {50\over 180}\pi \bigr )$ and

Now we have 5 equations and 5 unknowns

Then: solve

Post your work so we can comment and provide guidance

And here is why that's useful:

hy = 12 + x

First comment: be consistent with lower case/upper case and subscript. Not hx but H_x as in the sketch.

Learn a little $\LaTeX$
button at lower left:

Second: don't introduce a variable without clearly stating what it is. x ?

Third: Don't use a name that can lead to mistakes. x ?

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[hello478]

will solve it with numbers in a minute

 

[hello478]

Now we have 5 equations and 5 unknowns

ok, so im afraid i cant solve this too, i tried using my brain but i just cant solve it, i think its because i havent attempted a problem like this before... so please help me... im sorry for not grasping it quickly...

 

[BvU]

ok, so im afraid i cant solve this too, i tried using my brain but i just cant solve it, i think its because i havent attempted a problem like this before... so please help me... im sorry for not grasping it quickly...

No need to apologize.

Just so we are on the same page, can you make explicit what you are trying to solve ?

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[hello478]

No need to apologize.

Just so we are on the same page, can you make explicit what you are trying to solve ?

i want to find out the reaction forces, horizontal and vertical on the hinge like you told, i just dont understand anything, i do understand that
the total vertical forces = W of beam + w of box + Ty + Hy = 0
and about the horizontal forces sum would also be 0 = Hx + Tx
but then how would tension be different if the w and W are same...?
ok so i dont understand anything more than that
i dont understand why would the tension change and i certainly cannot prove it through numbers
for box at point a and b like you said

 

[BvU]

Ok, let's address this question why $T$ is no longer 17 N when the box is somewhere else than at 0.35 m.

In (c) the weight of the beam was established by taking moments about point A :
$$\sum \tau = 0 \ \ \Leftrightarrow \ \ W\times 0.25 + 12\ {\sf N} \ \times {\color{red}{0.35}} - T\sin\Bigl ({50\over 180}\pi\Bigr ) \times 0.5 = 0 $$one equation with one unknown, from which follows $W = 9.25$ N

The position (distance from point A) of the block is prominent in the torque balance.

Now we are a bit further: $W$ is established and does not change when we move the block. The only quantity that can maintain the torque balance is $T_y$ -- and thereby $T$.

With the block at point A, the torque balance is $$\sum \tau = 0 \ \ \Leftrightarrow \ \ W\times 0.25 + 12\ {\sf N} \ \times 0.35 - T\sin\Bigl ({50\over 180}\pi\Bigr ) \times 0.5 = 0 $$now with W = 9.25 N so that $T = ...$

Can you fill in the remainder ?
Dinner time again !

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[haruspex]

the total vertical forces = W of beam + w of box + Ty + Hy = 0
and about the horizontal forces sum would also be 0 = Hx + Tx
but then how would tension be different if the w and W are same...?

You also know the relationship between Tx and Ty. But those three equations are not enough to determine Hy, Hx and Ty. There is a continuous range of solutions.
The reason is that those equations only say the mass centre of the system is not accelerating. To be static, there must also be no rotational acceleration. So your fourth equation must be a torque balance.
The numbers in that equation will determine Hy, Hx and Ty.

 

[hello478]

i have now come to the conclusion that this question is not made for me to solve... and that this is soo difficult
i have soo many questions about this problem
why would even the tension change when nothing is added or removed?
and why does tension change when the block is moved
and how can it possibly change?
and wouldnt the horizontal component of tension remain same always hence the vertical too
and i can solve it
i just tried but nothing makes sense
i have difficulty in understand this problem mathematically, please explain using some theory with it

 

[erobz]

why would even the tension change when nothing is added or removed?

Because the summation of the torques about the hinge changes.

and why does tension change when the block is moved

because the summation of the torques about the hinge changes.

and how can it possibly change?

Because the distance from the hinge to the mass changed. Hold a gallon of water at your waist, then raise it laterally. Which position is more difficult to maintain?

and wouldnt the horizontal component of tension remain same always hence the vertical too

The horizontal and vertical components of the tension force change in the same ratio ( $\frac{T_y}{T_x} = \tan \theta = constant$ ) for this problem.

If I replace the mass with that of a house, and it was placed anywhere but directly over the hinge (think levers) do you not think the tension in that rope would change?

i have difficulty in understand this problem mathematically, please explain using some theory with it

What do you think everyone has been doing?

 

[hello478]

What do you think everyone has been doing?

omggg wow thank you for making me realise what everyone has been doing, can't thank your kindness enough, your kindness just adds such a beautiful charm to your personality

 

[erobz]

omggg wow thank you for making me realise what everyone has been doing, can't thank your kindness enough, your kindness just adds such a beautiful charm to your personality

You're welcome!

 

[hello478]

Are my answers correct?

box At A:
T= 6.37
Ty= 4.625
Tx= 3.88
Hx= 3.88
Hy= 16.625

 

[erobz]

All kidding aside... Perhaps if you bothered to digest what I wrote before focusing on that last part of the reply ( you know...the stuff you need to learn), and what others have spent hours trying to explain ( for free) by asking thoughtful questions you would be getting more out of this? We are trying... Well...at least I was going to try, but now I think I'll pass.

 

[erobz]

Are my answers correct?

box At A:
T= 6.37
Ty= 4.625
Tx= 3.88
Hx= 3.88
Hy= 16.625

Check your own work.

Do they satisfy Newtons Laws?

$$ \sum F_y = 0 , ~\sum F_x = 0 , ~\sum M_A = 0 $$

 

[hello478]

All kidding aside... Perhaps if you bothered to digest what I wrote before focusing on that last part of the reply, and what others have spent hours trying to explain by asking thoughtful questions you would be getting more out of this? We are trying...

Yeah, keeping all kidding aside... i digested everyones answers and nourished my brain with it, they all were very delicious, your answer was very fibrous though, but i liked it... helped my digestion
and ig because of all the help i have attempted on the question
and im very thankful to everyones thoughtful questions and amazing help
thank you everyone for your generous efforts in helping my concepts (and giving me yummy food to eat and digest)!

 

[hello478]

Check your own work.

Do they satisfy Newtons Laws?

i tried to, and they sort of did, but i need someone else to confirm...
thank you for giving me the idea to check my own work you know, never thought of that before