A community for students.
Here's the question you clicked on:
 0 viewing
melmel
 one year ago
solve the pink one
melmel
 one year ago
solve the pink one

This Question is Closed

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Blue: 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.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1"How" meaning to take moments?

melmel
 one year ago
Best ResponseYou've already chosen the best response.0yes but i'm confuse >.<

melmel
 one year ago
Best ResponseYou've already chosen the best response.0roller gve vertical force to the wall so therefore

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1dw:1375617070605:dw

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1So you're good for both problems?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Keep going, you're on the right track. I have an appointment, so have a little time left.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Yes, what about the diagram?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1In the case of the ladder, there is a horizontal frictional force, F.dw:1375617403359:dw

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1But taking moments at A will ignore that.

melmel
 one year ago
Best ResponseYou've already chosen the best response.0the roller does not give horizontal force master this will give only a horizontal force if the roller

melmel
 one year ago
Best ResponseYou've already chosen the best response.0is parallel to the wall

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1dw:1375617764714:dw Would that help?

melmel
 one year ago
Best ResponseYou've already chosen the best response.0yes u are correct tension is there now can you teach me to take moment

melmel
 one year ago
Best ResponseYou've already chosen the best response.0where can i put my moment

melmel
 one year ago
Best ResponseYou've already chosen the best response.0basically i dont know where can i put the moment >.<

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1dw:1375617910870:dw Moments tend to turn the object (bar) that you have isolated as a "freebody". 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, anticlockwise is negative. So far so good?

melmel
 one year ago
Best ResponseYou've already chosen the best response.0yes i want to use the moment clock wise

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1When 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).

melmel
 one year ago
Best ResponseYou've already chosen the best response.0yes i familiar that Ra=mg

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Are 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.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Yep, 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.

melmel
 one year ago
Best ResponseYou've already chosen the best response.0thats all for the blue one?

melmel
 one year ago
Best ResponseYou've already chosen the best response.0i need also to solve the red one >.< dont go :)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1By 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.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Can you draw the FBD for the red one?

melmel
 one year ago
Best ResponseYou've already chosen the best response.0let me put the Rb and Ra and correct me if im wrong

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Here you know (again) that T=W=mg. dw:1375619330484:dw

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1and that Ra=Rb by equilbrium of horizontal and vertical forces, right?

melmel
 one year ago
Best ResponseYou've already chosen the best response.0where did tou get 10 and 5 the distance of the assume box is 12m

melmel
 one year ago
Best ResponseYou've already chosen the best response.0i think yes their are equal but im not sure

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1Oh, 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! :)