Computing Reaction Forces of Supports A & B with a 2 Degree Load

AI Thread Summary
To compute the reaction forces of supports A and B under a 2-degree load, one can apply the principle that a distributed load can be treated as a concentrated load acting at its centroid. This approach simplifies the calculations, allowing for easier determination of the reaction forces. Integration may be necessary to accurately account for the load distribution. Relevant formulas for calculating reaction forces can be found in structural analysis resources. Understanding these concepts is crucial for solving the problem effectively.
plucker_08
Messages
54
Reaction score
0
guys, how can u compute for the reaction forces of supports A & B w/ a 2 degree load...?

a link is provided below

thanks!

http://img49.imageshack.us/img49/7876/19uh.jpg
 
Last edited by a moderator:
Physics news on Phys.org
Isn't there some theorem that says a distributed load can be considered the same as a concentrated load acting at the centroid? I think some integration is in order.
 
there are formulas but i forgot them..
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top