A uniform 250 N ladder rests against a perfectly smooth wall, making a 35 degree angle with the wall. (a) find the normal forces that the wall and the floor exert on the ladder. B) what is the friction force on the ladder at the floor?
Try to sketch a diagram as you read through this post. An image is worth a thousand words.
There are four forces acting on the ladder of length L and making θ=35 degrees with the vertical smooth wall and on a horizontal rough floor.
1. the weight W=250N acting downwards through the middle of the ladder
2. vertical reaction on the floor acting upwards, equal to W.
These two forces form a couple of magnitude
W(L/2)sinθ
3. Normal reaction N on the wall (horizontal) at the top end of the ladder, equal in magnitude to
4. Frictional force F acting horizontally at the bottom of the ladder.
These two forces form another couple equal to FLcosθ.
Since the ladder is in equilibrium, the two couples must be equal, thus
W(L/2)sinθ = FLcosθ
From which we can solve for F
= (W/2)tanθ
after cancelling L on each side and replacing sinθ/cosθ by tanθ.
so it would be 125 (tan 35) = 87.5 Awesome that is what it says in the back of the book THANK YOU!
Glad that it works out well!
In case you're still on, just want to let you know that I finish for the night. I can answer any question you may have tomorrow, unless another teacher is still on tonight.
To find the normal forces exerted by the wall and the floor on the ladder, we need to analyze the forces acting on the ladder.
(a) Normal Force by the Wall:
Considering the forces acting in the vertical direction (y-direction), we have:
Sum of vertical forces = 0
The vertical forces acting on the ladder are:
1. The weight of the ladder, which acts downwards and has a magnitude of 250 N.
2. The normal force exerted by the wall on the ladder, which acts vertically upwards.
Since the ladder is in equilibrium, the sum of the vertical forces must be zero:
N_wall - 250 N = 0
N_wall = 250 N
Therefore, the normal force exerted by the wall on the ladder is 250 N.
Normal Force by the Floor:
Considering the forces acting in the horizontal direction (x-direction), we have:
Sum of horizontal forces = 0
The horizontal forces acting on the ladder are:
1. The horizontal component of the weight of the ladder, which acts towards the wall and has a magnitude of 250 N * sin(35°). (Note: sin(35°) is the ratio of the length of the opposite side to the hypotenuse in a right triangle)
2. The normal force exerted by the floor on the ladder, which acts horizontally in the opposite direction.
Since the ladder is in equilibrium, the sum of the horizontal forces must be zero:
N_floor - 250 N * sin(35°) = 0
N_floor = 250 N * sin(35°)
Therefore, the normal force exerted by the floor on the ladder is 250 N * sin(35°).
(b) Friction Force on the Ladder at the Floor:
The friction force exerted by the floor on the ladder will oppose the horizontal component of the weight.
The friction force (F_friction) can be calculated using the equation:
F_friction = coefficient of friction * N_floor
However, we need to know the coefficient of friction between the ladder and the floor. Without this information, it is not possible to determine the exact value of the friction force.
Note: The coefficient of friction is a dimensionless quantity that varies depending on the nature of the surfaces in contact. It should be provided in the problem statement or given in context.