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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?
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?

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ujjwal
 4 years ago
Best ResponseYou've already chosen the best response.0Do 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
 4 years ago
Best ResponseYou've already chosen the best response.0yes the question modified....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0guys i need help @experimentX @VincentLyon.Fr

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are you sure the 30° angle is with the floor and not with the wall?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since 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
 4 years ago
Best ResponseYou've already chosen the best response.0the 30degrees is with the floor

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, now, have you tried applying the method I described earlier?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no idea @VincentLyon.Fr

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you know how to express the moment (torque) of a force about a certain point?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep i do but get confused with the center of gravity constraint

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What do you mean "constraint"?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the fact that the center of gravity of the ladder is one third the way up from the bottom.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 3 years ago
Best ResponseYou've already chosen the best response.0pls i need help its not clear..since the lenght of AB is not given..aw do we work out moment about point A

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1340887287738:dw 30cos50 X 150 = R X bc

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@VincentLyon.Fr u dig me?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0according to me no matter where the center of gravity lies. horizontal force applied must be 150cos30' .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here is the answer to your problem (see attached file):

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks for the lecture @VincentLyon.Fr
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