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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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ujjwalBest 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..
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
yes the question modified....
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
guys i need help @experimentX @VincentLyon.Fr
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
Are you sure the 30° angle is with the floor and not with the wall?
 one year ago

VincentLyon.FrBest 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.
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
the 30degrees is with the floor
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
Ok, now, have you tried applying the method I described earlier?
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
no idea @VincentLyon.Fr
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
Do you know how to express the moment (torque) of a force about a certain point?
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
yep i do but get confused with the center of gravity constraint
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
What do you mean "constraint"?
 one year ago

woleraymondBest 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.
 one year ago

VincentLyon.FrBest 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.
 one year ago

woleraymondBest 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
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
dw:1340887287738:dw 30cos50 X 150 = R X bc
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
@VincentLyon.Fr u dig me?
 one year ago

theyatinBest 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' .
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.1
Here is the answer to your problem (see attached file):
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
Thanks for the lecture @VincentLyon.Fr
 one year ago
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