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?

Question

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?

ujjwal · Answer

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 · Answer

yes the question modified....

anonymous · Answer

guys i need help @experimentX  @Vincent-Lyon.Fr

vincent-lyon.fr · Answer

Are you sure the 30° angle is with the floor and not with the wall?

vincent-lyon.fr · Answer

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.