# Physics -equilibrium

• Feb 21st 2010, 01:04 PM
mj.alawami
Physics -equilibrium
How to solve this problem ?
Attachment 15529

Thank you
• Feb 21st 2010, 01:28 PM
Hello mj.alawami
Quote:

Originally Posted by mj.alawami
How to solve this problem ?
Attachment 15529

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?

• Feb 21st 2010, 01:41 PM
mj.alawami
Quote:

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?
• Feb 21st 2010, 05:43 PM
skeeter
Quote:

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

$R_B = 379g \, N$
• Feb 21st 2010, 11:12 PM
Hello mj.alawami
Quote:

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).

• Feb 21st 2010, 11:40 PM
mj.alawami
Quote:

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 ?
• Feb 22nd 2010, 02:52 AM
$450g+220g -R_A - R_B = 0$
(Some people denote this by $\sum F_y=0$.)
Substitute the value we've found for $R_B$ and solve for $R_A$.