Mass and Force distribution between two legs

[akhter900]

...o..M
.... /|\
.Theta 2 /..|.\.Theta 1
.../...v..\
...F2 /...F...\.F1
.../.....\
.../... \
-----------|---------ground
...|...p...|
...v..f2.....v..f1

Hi every one,
On the above diagram, M=10g is mounted on two fixed rods from the ground. So, the force acting on the point 'o' is F=10*9.8 = 98N. This force is going to distribute through the support as F1 and F2 and force on the ground for that two support will be f1 and f2. Here f1 is closer to p than f2. So, f1 should be grater than f2, ie. f1 > f2, is not it? Theta 1 = 30 degree and theta 2 = 45 degree.

To find out F1 and F2 I did something like,
F1 = F/Cos(30) = 113.16N
F2 = F/Cos(45) = 138.59N
Is it correct for force vector? I need to proof that F1 > F2.

I could not find out the distributed force F1 and F2 again f1 and f2 also. Is there anybody to give me some suggestions on this. I will be so grateful to you.

 

[tiny-tim]

welcome to pf!

hi akhter900! welcome to pf!

(have a theta: θ and a degree: ° and try using the X₂ tag just above the Reply box )

To find out F1 and F2 I did something like,
F1 = F/Cos(30) = 113.16N
F2 = F/Cos(45) = 138.59N

no, you can't do that, you can't split up the forces on the same body …

F₁ and F₂ are both acting on M, so they must both be in any equation for M

start again, doing components of all forces on M in the x and y directions (separately) ​

 

[akhter900]

Dear Tiny-tim,
Thanks a lot for the quick reply. What if I do the following way...

If I consider the above configuration as a straight line where M acting on its center of mass then it is possible to find out F1 and F2, isn't it?

...x2.....x1
...M2......M, F...M1
...O---------------o----------O
---|---------------------------|----Ground
...v..F1. ......v...F2

So, F = 98N. x1 = 2 and x2 = 4.
F = F1 + F2------------(1)
=> F1 = F-F2
F1*x1 = F2*x2---------(2)
=> (98-F2)2 = (F2*4)
=> (4*F2 + 2*F2) = 2*98
=> F2 = (2*98)/6 = 32.66N
Similarly, F1 = 65.33N

So, F1 > F2 Proved. and the friction for F1 is grater than friction for F2.

Is the procedure correct. Need your valuable suggestion.

 

[tiny-tim]

hi akhter900!

If I consider the above configuration as a straight line where M acting on its center of mass then it is possible to find out F1 and F2, isn't it?

...x2.....x1
...M2......M, F...M1
...O---------------o----------O
---|---------------------------|----Ground
...v..F1. ......v...F2

hmm … you seem to be laying the whole thing out on the ground …

there might be some merit in that, but i don't see it yet

So, F = 98N. x1 = 2 and x2 = 4.
F = F1 + F2------------(1)
=> F1 = F-F2
F1*x1 = F2*x2---------(2)

no … (1) should be the equation for the y-components, and (2) should be the equation for the x-components

try again ​

 

[akhter900]

Do I need to consider x component here? The bar is considered as rigid. The total mass is acting on the two points F1 and F2. based on that, F1 > F2. its proved.

If you have any other idea or technique, please tell me. I will be so much grateful to you.

 

[tiny-tim]

The bar is considered as rigid.

what bar is considered as rigid?

i see nothing about rigidity in the question or in the diagram …

unless the whole joint at M is rigid, yes you must consider the x components

 

[akhter900]

According to the figure...(just uploaded)

If I consider the bar AB is rigid then the force F applied on the point O will divide and act on the two points A and B. Am I right?

The x component for X1 or X2 makes 0 degree with the horizontal line, so Cos0 will be 1. So the x component will be only X1 and X2. Is it correct?

F1 and F2 can be found by the mass distribution law mentioned earlier. F1 is grater than F2, so the friction on the point B will be more than the other point A.

Is it okay? Your suggestion please...

 

[akhter900]

Would you please give me some links or documents about this type of problem? I want to study on...

 

[tiny-tim]

hi akhter900!

If I consider the bar AB is rigid then the force F applied on the point O will divide and act on the two points A and B. Am I right?

This is a completely different situation (and incidentally your F₁ and F₂ in the new diagram correspond to f₁ and f₂ in your original diagram, not to F₁ and F₂).

You must use the original diagram.

I do not understand why you will not try using the component method, as suggested.