Solving Equilibrium: 80N Ladder on Wall & Ground

[IBB]

Homework Statement

A)A ladder of weight 80N rests against a smooth wall at an angle of 29 degrees to the vertical with its end on a rough horizontal floor.If the ladder is in equillibrium calculate the reaction (magnitude and direction) of the ground and the wall.
Bi)a man of weight 80N climbs to the top,determine the force reaction of the ground and the wall
Bii)What is the minimum value for μ the coefficient of friction if the ladder does not slip.
All the examples i look at involve a length for the ladder and a figure for μ.

Homework Equations

the sum of all the forces in the x direction are equal to zero and likewise in the y direction.

The Attempt at a Solution

I would have thought that it would just be a opposing the weight of the ladder so reaction at the floor would just be 80N in a upward direction and the reaction of the wall would be the force of friction μ to the right.

 

[bigfooted]

The two end points of the ladder both have a reaction force in x and y direction, so you have 4 unknowns. The problem states that the wall is smooth, so the vertical force at the top point of the ladder is Fy=0. You still need another equation.
2. The sum of the moments is zero. The sum of the moments around the point where the ladder touches the floor will give you another equation.

 

[IBB]

thanks for your reply,i have calculated the vertical forces to be equal to 80/Cos29=91.47N
Horizontal 91.47*Sin29=44.35N
Do you know whether these are correct?
In regards to the equation are they Newtons second and third laws?

 

[haruspex]

i have calculated the vertical forces to be equal to 80/Cos29=91.47N
Horizontal 91.47*Sin29=44.35N
Do you know whether these are correct?

No, they're not. Please post your working.

 

[HalfLostAndSearching]

However, since the ladder is in equilibrium, the sum of all the forces in the x and y directions must be zero. This means that the reaction at the floor must not only balance the weight of the ladder, but also any additional forces acting on the ladder, such as the force exerted by the man climbing it.

To determine the reaction at the ground and the wall when the man is climbing, we can use the equations for equilibrium. In the x direction, the sum of the forces must be zero, so we have:

R_wall + F_friction = 0

In the y direction, the sum of the forces must also be zero, so we have:

R_floor - mg = 0

where R_wall is the reaction of the wall, F_friction is the force of friction, R_floor is the reaction of the floor, and mg is the weight of the man.

To find the minimum value for μ, we can use the equation for the maximum possible friction force:

F_friction = μR_floor

Substituting this into the equation for the sum of forces in the x direction, we get:

R_wall + μR_floor = 0

Solving for μ, we get:

μ = -R_wall/R_floor

Since we know that R_floor is equal to mg, we can substitute this into the equation to get:

μ = -R_wall/mg

To find the minimum value for μ, we need to find the maximum value for R_wall. This occurs when the ladder is about to slip, so we can use the equation for the maximum possible friction force again:

F_friction = μR_floor

Substituting this into the equation for the sum of forces in the y direction, we get:

R_floor - μR_floor = 0

Solving for R_floor, we get:

R_floor = μR_floor

Substituting this into the equation for μ, we get:

μ = -R_wall/R_floor = -R_wall/(μR_floor) = -1/μ

Therefore, the minimum value for μ is 1. This means that the coefficient of friction must be at least 1 for the ladder to not slip.