Solving Equilibrium: 80N Ladder on Wall & Ground

AI Thread Summary
The discussion focuses on solving equilibrium conditions for an 80N ladder leaning against a smooth wall at a 29-degree angle. Participants analyze the forces acting on the ladder, including the reactions at the ground and wall, and the need for additional equations to account for the unknowns. The calculations for vertical and horizontal forces are debated, with one participant questioning the accuracy of their results. The importance of applying the sum of moments around the ladder's base to derive further equations is emphasized. Overall, the thread highlights the complexities of equilibrium problems involving multiple forces and moments.
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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.
 
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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.
 
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?
 
IBB said:
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.
 
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