1. ## Physics -equilibrium

How to solve this problem ?

Thank you

2. Hello mj.alawami
Originally Posted by mj.alawami
How to solve this problem ?

Thank you
The beam is $8$ m in length; it is uniform. So its weight acts $4$ m from each end.

If the reaction forces at A and B are $R_A$ and $R_B$ kg-wt respectively, take moments about A:
$450 \times 4 + 220\times 5.6 -8R_B=0$
Solve this equation for $R_B$.

Then resolve vertically (or take moments about B) to find $R_A$.

Can you complete it now?

Hello mj.alawamiThe beam is $8$ m in length; it is uniform. So its weight acts $4$ m from each end.

If the reaction forces at A and B are $R_A$ and $R_B$ kg-wt respectively, take moments about A:
$450 \times 4 + 220\times 5.6 -8R_B=0$
Solve this equation for $R_B$.

Then resolve vertically (or take moments about B) to find $R_A$.

Can you complete it now?

So is the answer RB=379N , and should i take into consideration the gravitational attraction or no?

4. Originally Posted by mj.alawami
So is the answer RB=379N , and should i take into consideration the gravitational attraction or no?
$R_B = 379g \, N$

5. Hello mj.alawami
Originally Posted by mj.alawami
So is the answer RB=379N , and should i take into consideration the gravitational attraction or no?
If you are using Newtons as the unit of force, then, yes: multiply by $g\; (\approx 9.8)$, as skeeter has shown you. I was perhaps unwise to suggest a different unit (kg-wt or kg-force).

Hello mj.alawamiIf you are using Newtons as the unit of force, then, yes: multiply by $g\; (\approx 9.8)$, as skeeter has shown you. I was perhaps unwise to suggest a different unit (kg-wt or kg-force).

Will Ra and Rb have the same answer ?

7. Hello mj.alawami
Originally Posted by mj.alawami
Will Ra and Rb have the same answer ?
No. Let me re-phrase my first post, take the sum of the vertical forces and equate to zero:
$450g+220g -R_A - R_B = 0$
(Some people denote this by $\sum F_y=0$.)

Substitute the value we've found for $R_B$ and solve for $R_A$.