anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ujjwal
  • 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..
anonymous
  • anonymous
yes the question modified....
anonymous
  • anonymous
guys i need help @experimentX @Vincent-Lyon.Fr

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Are you sure the 30° angle is with the floor and not with the wall?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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.
anonymous
  • anonymous
the 30degrees is with the floor
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Ok, now, have you tried applying the method I described earlier?
anonymous
  • anonymous
no idea @Vincent-Lyon.Fr
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Do you know how to express the moment (torque) of a force about a certain point?
anonymous
  • anonymous
yep i do but get confused with the center of gravity constraint
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
What do you mean "constraint"?
anonymous
  • anonymous
the fact that the center of gravity of the ladder is one third the way up from the bottom.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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.
anonymous
  • 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
anonymous
  • anonymous
|dw:1340887287738:dw| 30cos50 X 150 = R X bc
anonymous
  • anonymous
@Vincent-Lyon.Fr u dig me?
anonymous
  • anonymous
according to me no matter where the center of gravity lies. horizontal force applied must be 150cos30' .
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Here is the answer to your problem (see attached file):
1 Attachment
anonymous
  • anonymous
Thanks for the lecture @Vincent-Lyon.Fr

Looking for something else?

Not the answer you are looking for? Search for more explanations.