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