## anonymous 4 years ago a 150kg ladder leans against a smooth wall,making an angle 30 degrees with the floor.The center of gravity of the ladder is one third the way up from the bottom. How large a horizontal force must the floor provide if the ladder is not to slip?

1. ujjwal

Do you have a diagram? the floor provides horizontal force to what? the ladder or wall.. I guess you are talking about ladder.. But are you sure it is the center of gravity of wall whose position is given? I guess it should be of ladder..

2. anonymous

yes the question modified....

3. anonymous

guys i need help @experimentX @Vincent-Lyon.Fr

4. anonymous

Are you sure the 30° angle is with the floor and not with the wall?

5. anonymous

Since the ladder is in equilibrium, net torque about any point must be zero. Chose point A where ladder is touching the floor.|dw:1337890332208:dw| It will allow you to find the value of $$N_B$$. Then write that net force is also zero. Finding horizontal component of $$\vec R_A$$ is obvious.

6. anonymous

the 30degrees is with the floor

7. anonymous

Ok, now, have you tried applying the method I described earlier?

8. anonymous

no idea @Vincent-Lyon.Fr

9. anonymous

Do you know how to express the moment (torque) of a force about a certain point?

10. anonymous

yep i do but get confused with the center of gravity constraint

11. anonymous

What do you mean "constraint"?

12. anonymous

the fact that the center of gravity of the ladder is one third the way up from the bottom.

13. anonymous

Well, as you know where it is, you know its distance from A is l/3. That enables you to work out moment of weight about point A.

14. anonymous

pls i need help its not clear..since the lenght of AB is not given..aw do we work out moment about point A

15. anonymous

|dw:1340887287738:dw| 30cos50 X 150 = R X bc

16. anonymous

@Vincent-Lyon.Fr u dig me?

17. anonymous

according to me no matter where the center of gravity lies. horizontal force applied must be 150cos30' .

18. anonymous