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woleraymond
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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?
 2 years ago
 2 years ago
woleraymond Group Title
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?
 2 years ago
 2 years ago

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ujjwal Group TitleBest ResponseYou've already chosen the best response.0
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 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
yes the question modified....
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
guys i need help @experimentX @VincentLyon.Fr
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
Are you sure the 30° angle is with the floor and not with the wall?
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
the 30degrees is with the floor
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
Ok, now, have you tried applying the method I described earlier?
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
no idea @VincentLyon.Fr
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
Do you know how to express the moment (torque) of a force about a certain point?
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
yep i do but get confused with the center of gravity constraint
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
What do you mean "constraint"?
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
the fact that the center of gravity of the ladder is one third the way up from the bottom.
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
pls i need help its not clear..since the lenght of AB is not given..aw do we work out moment about point A
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
dw:1340887287738:dw 30cos50 X 150 = R X bc
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@VincentLyon.Fr u dig me?
 2 years ago

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

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
Here is the answer to your problem (see attached file):
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
Thanks for the lecture @VincentLyon.Fr
 2 years ago
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