A 150-N ladder leans against a smooth wall, inclined at an angle of 60 degrees above the floor. The centre of gravity of the ladder is at point on the ladder which is one-third the length of the ladder from the bottom. How large a horizontal force must the floor provide if the ladder is not to slip?
I wish i could post a link of my FBD here, but the website won't allow any links to be posted.
It's really hard to explain on here if you can't see a free body diagram.
Could you show me how you started it and maybe I can try to follow what you're doing?
If you want to check your answer, I got 283.19N
To find the horizontal force required to prevent the ladder from slipping, we need to consider the forces acting on the ladder.
1. Weight of the ladder: The ladder weighs 150 N, which can be considered as acting downward through its center of gravity.
2. Normal force: The smooth wall exerts a normal force perpendicular to the wall, effectively canceling out the vertical component of the weight.
3. Friction force: The floor must exert a horizontal force to counteract the tendency of the ladder to slip.
Let's determine the forces acting on the ladder and the required horizontal force.
1. Weight component: We can split the weight force into two components: vertical and horizontal. The vertical component is given by: weight * sin(60°).
Weight vertical component = 150 N * sin(60°) = 150 N * 0.866 = 129.9 N.
2. Normal force: The normal force cancels out the vertical component of the weight, so the normal force is equal to the weight vertical component.
Normal force = 129.9 N.
3. Friction force: The friction force opposes the horizontal component of the weight.
Friction force = weight * cos(60°) = 150 N * cos(60°) = 150 N * 0.5 = 75 N.
The floor must provide a horizontal force equal to the friction force to prevent the ladder from slipping. Therefore, the required horizontal force is 75 N.