Here's the question you clicked on:
melmel
solve the pink one
Blue: Denote tension in cord as T. Vertical reaction on bar equals weight of bar, assuming no friction anywhere. Take moments about B. Solve for T.
"How" meaning to take moments?
roller gve vertical force to the wall so therefore
I can give an example: |dw:1375616654183:dw| A ladder 5 m long leans agains a smooth wall at B and stands on a rough floor at A at a distance of 3 m from the wall as shown above. The mass of the ladder is 10 kg. Since we don't know the friction on the floor, we can take moments about A so the friction does not come in the equation. Let reaction at B = R Mass at the middle (C) = m Take moments about A, since the ladder is in equilibrium, sum of moments = 0. -R*4 + mg*(3/2)=0 Note: moment = force * distance, clockwise is positive. Solve for R: R=(3mg/2)/4, or =3mg/8 N.
So you're good for both problems?
Keep going, you're on the right track. I have an appointment, so have a little time left.
Yes, what about the diagram?
In the case of the ladder, there is a horizontal frictional force, F.|dw:1375617403359:dw|
But taking moments at A will ignore that.
the roller does not give horizontal force master this will give only a horizontal force if the roller
|dw:1375617764714:dw| Would that help?
yes u are correct tension is there now can you teach me to take moment
where can i put my moment
basically i dont know where can i put the moment >.<
|dw:1375617910870:dw| Moments tend to turn the object (bar) that you have isolated as a "free-body". Moment means the product of a force times the distance about a point. The distance must be the perpendicular distance, i.e. the shortest possible distance between the point and the direction of the force. Moments that turn the (free) body in a clockwise direction is positive, anti-clockwise is negative. So far so good?
yes i want to use the moment clock wise
When you take moments about a point (say B), we assume that B does not move but free to turn. When the (free) body is in equilibrium, the sum of moments about any point is zero. So take moments about B: Ra*(3sin(60) - T(2cos(60)) = 0 Since Rb passes through point B, the distance is zero, so the product (moment) is also zero. Now you can solve for T in terms of Ra. (You know from the sum of forces in the vertical direction that Ra=mg).
yes i familiar that Ra=mg
Are you able to solve for T now? By the way, in the future you would get more attention posting mechanics problems under Physics or Engineering. I enjoy solving mechanics problems.
Yep, that's what I got (1274.356). You can solve the blue problem the same way. In any case, I've got to go. If you need more help, you may want to post the question again under physics or engineering, whichever you're onto.
thats all for the blue one?
i need also to solve the red one >.< dont go :)
By the way, the minor difference is because I use g=9.81, when most people use 9.8. Yes, that's all for the blue one. Sorry, I meant you can solve the red one the same way.
Can you draw the FBD for the red one?
let me put the Rb and Ra and correct me if im wrong
Here you know (again) that T=W=mg. |dw:1375619330484:dw|
and that Ra=Rb by equilbrium of horizontal and vertical forces, right?
where did tou get 10 and 5 the distance of the assume box is 12m
i think yes their are equal but im not sure
Oh, I was sloppy. It should have been 4 and 8 to make 12. The answer would have been the same. (you don't need to know that! :)